def limit(e, z, z0, dir="+"):
"""
Compute the limit of e(z) at the point z0.
z0 can be any expression, including oo and -oo.
For dir="+" (default) it calculates the limit from the right
(z->z0+) and for dir="-" the limit from the left (z->z0-). For infinite z0
(oo or -oo), the dir argument doesn't matter.
Examples
========
>>> from sympy import limit, sin, Symbol, oo
>>> from sympy.abc import x
>>> limit(sin(x)/x, x, 0)
1
>>> limit(1/x, x, 0, dir="+")
oo
>>> limit(1/x, x, 0, dir="-")
-oo
>>> limit(1/x, x, oo)
0
Notes
=====
First we try some heuristics for easy and frequent cases like "x", "1/x",
"x**2" and similar, so that it's fast. For all other cases, we use the
Gruntz algorithm (see the gruntz() function).
"""
from sympy import Wild, log
e = sympify(e)
z = sympify(z)
z0 = sympify(z0)
if e == z:
return z0
if e.is_Rational:
return e
if not e.has(z):
return e
if e.func is tan:
# discontinuity at odd multiples of pi/2; 0 at even
disc = S.Pi / 2
sign = 1
if dir == '-':
sign *= -1
i = limit(sign * e.args[0], z, z0) / disc
if i.is_integer:
if i.is_even:
return S.Zero
elif i.is_odd:
if dir == '+':
return S.NegativeInfinity
else:
return S.Infinity
if e.func is cot:
# discontinuity at multiples of pi; 0 at odd pi/2 multiples
disc = S.Pi
sign = 1
if dir == '-':
sign *= -1
i = limit(sign * e.args[0], z, z0) / disc
if i.is_integer:
if dir == '-':
return S.NegativeInfinity
else:
return S.Infinity
elif (2 * i).is_integer:
return S.Zero
if e.is_Pow:
b, ex = e.args
c = None # records sign of b if b is +/-z or has a bounded value
if b.is_Mul:
c, b = b.as_two_terms()
if c is S.NegativeOne and b == z:
c = '-'
elif b == z:
c = '+'
if ex.is_number:
if c is None:
base = b.subs(z, z0)
if base.is_bounded and (ex.is_bounded or base is not S.One):
return base**ex
else:
if z0 == 0 and ex < 0:
if dir != c:
# integer
if ex.is_even:
return S.Infinity
elif ex.is_odd:
return S.NegativeInfinity
# rational
elif ex.is_Rational:
return (S.NegativeOne**ex) * S.Infinity
else:
return S.ComplexInfinity
return S.Infinity
return z0**ex
if e.is_Mul or not z0 and e.is_Pow and b.func is log:
if e.is_Mul:
# weed out the z-independent terms
i, d = e.as_independent(z)
if i is not S.One and i.is_bounded:
return i * limit(d, z, z0, dir)
else:
i, d = S.One, e
if not z0:
# look for log(z)**q or z**p*log(z)**q
p, q = Wild("p"), Wild("q")
r = d.match(z**p * log(z)**q)
if r:
p, q = [r.get(w, w) for w in [p, q]]
if q and q.is_number and p.is_number:
if q > 0:
if p > 0:
return S.Zero
else:
return -oo * i
else:
if p >= 0:
return S.Zero
else:
return -oo * i
if e.is_Add:
if e.is_polynomial() and not z0.is_unbounded:
return Add(*[limit(term, z, z0, dir) for term in e.args])
# this is a case like limit(x*y+x*z, z, 2) == x*y+2*x
# but we need to make sure, that the general gruntz() algorithm is
# executed for a case like "limit(sqrt(x+1)-sqrt(x),x,oo)==0"
unbounded = []
unbounded_result = []
finite = []
unknown = []
ok = True
for term in e.args:
if not term.has(z) and not term.is_unbounded:
finite.append(term)
continue
result = term.subs(z, z0)
bounded = result.is_bounded
if bounded is False or result is S.NaN:
if unknown:
ok = False
break
unbounded.append(term)
if result != S.NaN:
# take result from direction given
result = limit(term, z, z0, dir)
unbounded_result.append(result)
elif bounded:
finite.append(result)
else:
if unbounded:
ok = False
break
unknown.append(result)
if not ok:
# we won't be able to resolve this with unbounded
# terms, e.g. Sum(1/k, (k, 1, n)) - log(n) as n -> oo:
# since the Sum is unevaluated it's boundedness is
# unknown and the log(n) is oo so you get Sum - oo
# which is unsatisfactory.
raise NotImplementedError('unknown boundedness for %s' %
(unknown or result))
u = Add(*unknown)
if unbounded:
inf_limit = Add(*unbounded_result)
if inf_limit is not S.NaN:
return inf_limit + u
if finite:
return Add(*finite) + limit(Add(*unbounded), z, z0, dir) + u
else:
return Add(*finite) + u
if e.is_Order:
args = e.args
return C.Order(limit(args[0], z, z0), *args[1:])
try:
r = gruntz(e, z, z0, dir)
if r is S.NaN:
raise PoleError()
except PoleError:
r = heuristics(e, z, z0, dir)
return r
def limit(e, z, z0, dir="+"):
"""
Compute the limit of e(z) at the point z0.
z0 can be any expression, including oo and -oo.
For dir="+" (default) it calculates the limit from the right
(z->z0+) and for dir="-" the limit from the left (z->z0-). For infinite z0
(oo or -oo), the dir argument doesn't matter.
Examples
========
>>> from sympy import limit, sin, Symbol, oo
>>> from sympy.abc import x
>>> limit(sin(x)/x, x, 0)
1
>>> limit(1/x, x, 0, dir="+")
oo
>>> limit(1/x, x, 0, dir="-")
-oo
>>> limit(1/x, x, oo)
0
Notes
=====
First we try some heuristics for easy and frequent cases like "x", "1/x",
"x**2" and similar, so that it's fast. For all other cases, we use the
Gruntz algorithm (see the gruntz() function).
"""
from sympy import Wild, log
e = sympify(e)
z = sympify(z)
z0 = sympify(z0)
if e == z:
return z0
if e.is_Rational:
return e
if not e.has(z):
return e
# gruntz fails on factorials but works with the gamma function
# If no factorial term is present, e should remain unchanged.
# factorial is defined to be zero for negative inputs (which
# differs from gamma) so only rewrite for positive z0.
if z0.is_positive:
e = e.rewrite(factorial, gamma)
if e.func is tan:
# discontinuity at odd multiples of pi/2; 0 at even
disc = S.Pi/2
sign = 1
if dir == '-':
sign *= -1
i = limit(sign*e.args[0], z, z0)/disc
if i.is_integer:
if i.is_even:
return S.Zero
elif i.is_odd:
if dir == '+':
return S.NegativeInfinity
else:
return S.Infinity
if e.func is cot:
# discontinuity at multiples of pi; 0 at odd pi/2 multiples
disc = S.Pi
sign = 1
if dir == '-':
sign *= -1
i = limit(sign*e.args[0], z, z0)/disc
if i.is_integer:
if dir == '-':
return S.NegativeInfinity
else:
return S.Infinity
elif (2*i).is_integer:
return S.Zero
if e.is_Pow:
b, ex = e.args
c = None # records sign of b if b is +/-z or has a bounded value
if b.is_Mul:
c, b = b.as_two_terms()
if c is S.NegativeOne and b == z:
c = '-'
elif b == z:
c = '+'
if ex.is_number:
if c is None:
base = b.subs(z, z0)
if base.is_finite and (ex.is_bounded or base is not S.One):
return base**ex
else:
if z0 == 0 and ex < 0:
if dir != c:
# integer
if ex.is_even:
return S.Infinity
elif ex.is_odd:
return S.NegativeInfinity
# rational
elif ex.is_Rational:
return (S.NegativeOne**ex)*S.Infinity
else:
return S.ComplexInfinity
return S.Infinity
return z0**ex
if e.is_Mul or not z0 and e.is_Pow and b.func is log:
if e.is_Mul:
if abs(z0) is S.Infinity:
n, d = e.as_numer_denom()
# XXX todo: this should probably be stated in the
# negative -- i.e. to exclude expressions that should
# not be handled this way but I'm not sure what that
# condition is; when ok is True it means that the leading
# term approach is going to succeed (hopefully)
ok = lambda w: (z in w.free_symbols and
any(a.is_polynomial(z) or
any(z in m.free_symbols and m.is_polynomial(z)
for m in Mul.make_args(a))
for a in Add.make_args(w)))
if all(ok(w) for w in (n, d)):
u = C.Dummy(positive=(z0 is S.Infinity))
inve = (n/d).subs(z, 1/u)
return limit(inve.as_leading_term(u), u,
S.Zero, "+" if z0 is S.Infinity else "-")
# weed out the z-independent terms
i, d = e.as_independent(z)
if i is not S.One and i.is_bounded:
return i*limit(d, z, z0, dir)
else:
i, d = S.One, e
if not z0:
# look for log(z)**q or z**p*log(z)**q
p, q = Wild("p"), Wild("q")
r = d.match(z**p * log(z)**q)
if r:
p, q = [r.get(w, w) for w in [p, q]]
if q and q.is_number and p.is_number:
if q > 0:
if p > 0:
return S.Zero
else:
return -oo*i
else:
if p >= 0:
return S.Zero
else:
return -oo*i
if e.is_Add:
if e.is_polynomial():
if not z0.is_unbounded:
return Add(*[limit(term, z, z0, dir) for term in e.args])
elif e.is_rational_function(z):
rval = Add(*[limit(term, z, z0, dir) for term in e.args])
if rval != S.NaN:
return rval
if not any([a.is_unbounded for a in e.args]):
e = e.normal() # workaround for issue 3744
if e.is_Order:
args = e.args
return C.Order(limit(args[0], z, z0), *args[1:])
try:
r = gruntz(e, z, z0, dir)
if r is S.NaN:
raise PoleError()
except (PoleError, ValueError):
r = heuristics(e, z, z0, dir)
return r
def limit(e, z, z0, dir="+"):
"""
Compute the limit of e(z) at the point z0.
z0 can be any expression, including oo and -oo.
For dir="+" (default) it calculates the limit from the right
(z->z0+) and for dir="-" the limit from the left (z->z0-). For infinite z0
(oo or -oo), the dir argument doesn't matter.
Examples
========
>>> from sympy import limit, sin, Symbol, oo
>>> from sympy.abc import x
>>> limit(sin(x)/x, x, 0)
1
>>> limit(1/x, x, 0, dir="+")
oo
>>> limit(1/x, x, 0, dir="-")
-oo
>>> limit(1/x, x, oo)
0
Notes
=====
First we try some heuristics for easy and frequent cases like "x", "1/x",
"x**2" and similar, so that it's fast. For all other cases, we use the
Gruntz algorithm (see the gruntz() function).
"""
from sympy import Wild, log
e = sympify(e)
z = sympify(z)
z0 = sympify(z0)
if e == z:
return z0
if e.is_Rational:
return e
if not e.has(z):
return e
if e.func is tan:
# discontinuity at odd multiples of pi/2; 0 at even
disc = S.Pi / 2
sign = 1
if dir == "-":
sign *= -1
i = limit(sign * e.args[0], z, z0) / disc
if i.is_integer:
if i.is_even:
return S.Zero
elif i.is_odd:
if dir == "+":
return S.NegativeInfinity
else:
return S.Infinity
if e.func is cot:
# discontinuity at multiples of pi; 0 at odd pi/2 multiples
disc = S.Pi
sign = 1
if dir == "-":
sign *= -1
i = limit(sign * e.args[0], z, z0) / disc
if i.is_integer:
if dir == "-":
return S.NegativeInfinity
else:
return S.Infinity
elif (2 * i).is_integer:
return S.Zero
if e.is_Pow:
b, ex = e.args
c = None # records sign of b if b is +/-z or has a bounded value
if b.is_Mul:
c, b = b.as_two_terms()
if c is S.NegativeOne and b == z:
c = "-"
elif b == z:
c = "+"
if ex.is_number:
if c is None:
base = b.subs(z, z0)
if base.is_bounded and (ex.is_bounded or base is not S.One):
return base ** ex
else:
if z0 == 0 and ex < 0:
if dir != c:
# integer
if ex.is_even:
return S.Infinity
elif ex.is_odd:
return S.NegativeInfinity
# rational
elif ex.is_Rational:
return (S.NegativeOne ** ex) * S.Infinity
else:
return S.ComplexInfinity
return S.Infinity
return z0 ** ex
if e.is_Mul or not z0 and e.is_Pow and b.func is log:
if e.is_Mul:
# weed out the z-independent terms
i, d = e.as_independent(z)
if i is not S.One and i.is_bounded:
return i * limit(d, z, z0, dir)
else:
i, d = S.One, e
if not z0:
# look for log(z)**q or z**p*log(z)**q
p, q = Wild("p"), Wild("q")
r = d.match(z ** p * log(z) ** q)
if r:
p, q = [r.get(w, w) for w in [p, q]]
if q and q.is_number and p.is_number:
if q > 0:
if p > 0:
return S.Zero
else:
return -oo * i
else:
if p >= 0:
return S.Zero
else:
return -oo * i
if e.is_Add:
if e.is_polynomial() and not z0.is_unbounded:
return Add(*[limit(term, z, z0, dir) for term in e.args])
# this is a case like limit(x*y+x*z, z, 2) == x*y+2*x
# but we need to make sure, that the general gruntz() algorithm is
# executed for a case like "limit(sqrt(x+1)-sqrt(x),x,oo)==0"
unbounded = []
unbounded_result = []
finite = []
unknown = []
ok = True
for term in e.args:
if not term.has(z) and not term.is_unbounded:
finite.append(term)
continue
result = term.subs(z, z0)
bounded = result.is_bounded
if bounded is False or result is S.NaN:
if unknown:
ok = False
break
unbounded.append(term)
if result != S.NaN:
# take result from direction given
result = limit(term, z, z0, dir)
unbounded_result.append(result)
elif bounded:
finite.append(result)
else:
if unbounded:
ok = False
break
unknown.append(result)
if not ok:
# we won't be able to resolve this with unbounded
# terms, e.g. Sum(1/k, (k, 1, n)) - log(n) as n -> oo:
# since the Sum is unevaluated it's boundedness is
# unknown and the log(n) is oo so you get Sum - oo
# which is unsatisfactory.
raise NotImplementedError("unknown boundedness for %s" % (unknown or result))
u = Add(*unknown)
if unbounded:
inf_limit = Add(*unbounded_result)
if inf_limit is not S.NaN:
return inf_limit + u
if finite:
return Add(*finite) + limit(Add(*unbounded), z, z0, dir) + u
else:
return Add(*finite) + u
if e.is_Order:
args = e.args
return C.Order(limit(args[0], z, z0), *args[1:])
try:
r = gruntz(e, z, z0, dir)
if r is S.NaN:
raise PoleError()
except PoleError:
r = heuristics(e, z, z0, dir)
return r
def limit(e, z, z0, dir="+"):
"""
Compute the limit of e(z) at the point z0.
z0 can be any expression, including oo and -oo.
For dir="+" (default) it calculates the limit from the right
(z->z0+) and for dir="-" the limit from the left (z->z0-). For infinite z0
(oo or -oo), the dir argument doesn't matter.
Examples:
>>> from sympy import limit, sin, Symbol, oo
>>> from sympy.abc import x
>>> limit(sin(x)/x, x, 0)
1
>>> limit(1/x, x, 0, dir="+")
oo
>>> limit(1/x, x, 0, dir="-")
-oo
>>> limit(1/x, x, oo)
0
Strategy:
First we try some heuristics for easy and frequent cases like "x", "1/x",
"x**2" and similar, so that it's fast. For all other cases, we use the
Gruntz algorithm (see the gruntz() function).
"""
e = sympify(e)
z = sympify(z)
z0 = sympify(z0)
if e == z:
return z0
if e.is_Rational:
return e
if e.is_Pow and e.args[0] == z and e.args[1].is_number:
if e.args[1] > 0:
return z0**e.args[1]
if z0 == 0:
if dir == "+":
return S.Infinity
return -S.Infinity
return z0**e.args[1]
if e.is_Add:
if e.is_polynomial() and z0.is_finite:
return Add(*[limit(term, z, z0, dir) for term in e.args])
# this is a case like limit(x*y+x*z, z, 2) == x*y+2*x
# but we need to make sure, that the general gruntz() algorithm is
# executed for a case like "limit(sqrt(x+1)-sqrt(x),x,oo)==0"
unbounded = []; unbounded_result=[]
finite = []
for term in e.args:
result = term.subs(z, z0)
if result.is_unbounded or result is S.NaN:
unbounded.append(term)
unbounded_result.append(result)
else:
finite.append(result)
if unbounded:
inf_limit = Add(*unbounded_result)
if inf_limit is not S.NaN:
return inf_limit
if finite:
return Add(*finite) + limit(Add(*unbounded), z, z0, dir)
else:
return Add(*finite)
try:
r = gruntz(e, z, z0, dir)
except PoleError:
r = heuristics(e, z, z0, dir)
return r
def limit(e, z, z0, dir="+"):
"""
Compute the limit of e(z) at the point z0.
z0 can be any expression, including oo and -oo.
For dir="+" (default) it calculates the limit from the right
(z->z0+) and for dir="-" the limit from the left (z->z0-). For infinite z0
(oo or -oo), the dir argument doesn't matter.
Examples
========
>>> from sympy import limit, sin, Symbol, oo
>>> from sympy.abc import x
>>> limit(sin(x)/x, x, 0)
1
>>> limit(1/x, x, 0, dir="+")
oo
>>> limit(1/x, x, 0, dir="-")
-oo
>>> limit(1/x, x, oo)
0
Notes
=====
First we try some heuristics for easy and frequent cases like "x", "1/x",
"x**2" and similar, so that it's fast. For all other cases, we use the
Gruntz algorithm (see the gruntz() function).
"""
from sympy import Wild, log
e = sympify(e)
z = sympify(z)
z0 = sympify(z0)
if e == z:
return z0
if e.is_Rational:
return e
if not e.has(z):
return e
# gruntz fails on factorials but works with the gamma function
# If no factorial term is present, e should remain unchanged.
# factorial is defined to be zero for negative inputs (which
# differs from gamma) so only rewrite for positive z0.
if z0.is_positive:
e = e.rewrite(factorial, gamma)
if e.func is tan:
# discontinuity at odd multiples of pi/2; 0 at even
disc = S.Pi/2
sign = 1
if dir == '-':
sign *= -1
i = limit(sign*e.args[0], z, z0)/disc
if i.is_integer:
if i.is_even:
return S.Zero
elif i.is_odd:
if dir == '+':
return S.NegativeInfinity
else:
return S.Infinity
if e.func is cot:
# discontinuity at multiples of pi; 0 at odd pi/2 multiples
disc = S.Pi
sign = 1
if dir == '-':
sign *= -1
i = limit(sign*e.args[0], z, z0)/disc
if i.is_integer:
if dir == '-':
return S.NegativeInfinity
else:
return S.Infinity
elif (2*i).is_integer:
return S.Zero
if e.is_Pow:
b, ex = e.args
c = None # records sign of b if b is +/-z or has a bounded value
if b.is_Mul:
c, b = b.as_two_terms()
if c is S.NegativeOne and b == z:
c = '-'
elif b == z:
c = '+'
if ex.is_number:
if c is None:
base = b.subs(z, z0)
if base.is_finite and (ex.is_bounded or base is not S.One):
return base**ex
else:
if z0 == 0 and ex < 0:
if dir != c:
# integer
if ex.is_even:
return S.Infinity
elif ex.is_odd:
return S.NegativeInfinity
# rational
elif ex.is_Rational:
return (S.NegativeOne**ex)*S.Infinity
else:
return S.ComplexInfinity
return S.Infinity
return z0**ex
if e.is_Mul or not z0 and e.is_Pow and b.func is log:
if e.is_Mul:
if abs(z0) is S.Infinity:
n, d = e.as_numer_denom()
# XXX todo: this should probably be stated in the
# negative -- i.e. to exclude expressions that should
# not be handled this way but I'm not sure what that
# condition is; when ok is True it means that the leading
# term approach is going to succeed (hopefully)
ok = lambda w: (z in w.free_symbols and
any(a.is_polynomial(z) or
any(z in m.free_symbols and m.is_polynomial(z)
for m in Mul.make_args(a))
for a in Add.make_args(w)))
if all(ok(w) for w in (n, d)):
u = C.Dummy(positive=(z0 is S.Infinity))
inve = (n/d).subs(z, 1/u)
return limit(inve.as_leading_term(u), u,
S.Zero, "+" if z0 is S.Infinity else "-")
# weed out the z-independent terms
i, d = e.as_independent(z)
if i is not S.One and i.is_bounded:
return i*limit(d, z, z0, dir)
else:
i, d = S.One, e
if not z0:
# look for log(z)**q or z**p*log(z)**q
p, q = Wild("p"), Wild("q")
r = d.match(z**p * log(z)**q)
if r:
p, q = [r.get(w, w) for w in [p, q]]
if q and q.is_number and p.is_number:
if q > 0:
if p > 0:
return S.Zero
else:
return -oo*i
else:
if p >= 0:
return S.Zero
else:
return -oo*i
if e.is_Add:
if e.is_polynomial() and not z0.is_unbounded:
return Add(*[limit(term, z, z0, dir) for term in e.args])
# this is a case like limit(x*y+x*z, z, 2) == x*y+2*x
# but we need to make sure, that the general gruntz() algorithm is
# executed for a case like "limit(sqrt(x+1)-sqrt(x),x,oo)==0"
unbounded = []
unbounded_result = []
unbounded_const = []
unknown = []
unknown_result = []
finite = []
zero = []
def _sift(term):
if z not in term.free_symbols:
if term.is_unbounded:
unbounded_const.append(term)
else:
finite.append(term)
else:
result = term.subs(z, z0)
bounded = result.is_bounded
if bounded is False or result is S.NaN:
unbounded.append(term)
if result != S.NaN:
# take result from direction given
result = limit(term, z, z0, dir)
unbounded_result.append(result)
elif bounded:
if result:
finite.append(result)
else:
zero.append(term)
else:
unknown.append(term)
unknown_result.append(result)
for term in e.args:
_sift(term)
bad = bool(unknown and unbounded)
if bad or len(unknown) > 1 or len(unbounded) > 1 and not zero:
uu = unknown + unbounded
# we won't be able to resolve this with unbounded
# terms, e.g. Sum(1/k, (k, 1, n)) - log(n) as n -> oo:
# since the Sum is unevaluated it's boundedness is
# unknown and the log(n) is oo so you get Sum - oo
# which is unsatisfactory. BUT...if there are both
# unknown and unbounded terms (condition 'bad') or
# there are multiple terms that are unknown, or
# there are multiple symbolic unbounded terms they may
# respond better if they are made into a rational
# function, so give them a chance to do so before
# reporting failure.
u = Add(*uu)
f = u.normal()
if f != u:
unknown = []
unbounded = []
unbounded_result = []
unknown_result = []
_sift(limit(f, z, z0, dir))
# We came in with a) unknown and unbounded terms or b) had multiple
# unknown terms
# At this point we've done one of 3 things.
# (1) We did nothing with f so we now report the error
# showing the troublesome terms which are now in uu. OR
# (2) We did something with f but the result came back as unknown.
# Normally this wouldn't be a problem,
# but we had either multiple terms that were troublesome (unk and
# unbounded or multiple unknown terms) so if we
# weren't able to resolve the boundedness by now, that indicates a
# problem so we report the error showing the troublesome terms which are
# now in uu.
if unknown:
if bad:
msg = 'unknown and unbounded terms present in %s'
elif unknown:
msg = 'multiple terms with unknown boundedness in %s'
raise NotImplementedError(msg % uu)
# OR
# (3) the troublesome terms have been identified as finite or unbounded
# and we proceed with the non-error code since the lists have been updated.
u = Add(*unknown_result)
if unbounded_result or unbounded_const:
unbounded.extend(zero)
inf_limit = Add(*(unbounded_result + unbounded_const))
if inf_limit is not S.NaN:
return inf_limit + u
if finite:
return Add(*finite) + limit(Add(*unbounded), z, z0, dir) + u
else:
return Add(*finite) + u
if e.is_Order:
args = e.args
return C.Order(limit(args[0], z, z0), *args[1:])
try:
r = gruntz(e, z, z0, dir)
if r is S.NaN:
raise PoleError()
except (PoleError, ValueError):
r = heuristics(e, z, z0, dir)
return r
def limit(e, z, z0, dir="+"):
"""
Compute the limit of e(z) at the point z0.
z0 can be any expression, including oo and -oo.
For dir="+" (default) it calculates the limit from the right
(z->z0+) and for dir="-" the limit from the left (z->z0-). For infinite z0
(oo or -oo), the dir argument doesn't matter.
Examples:
>>> from sympy import limit, sin, Symbol
>>> x = Symbol('x')
>>> limit(sin(x)/x, x, 0)
1
>>> limit(1/x, x, 0, dir="+")
oo
>>> limit(1/x, x, 0, dir="-")
-oo
>>> limit(1/x, x, oo)
0
Strategy:
First we try some heuristics for easy and frequent cases like "x", "1/x",
"x**2" and similar, so that it's fast. For all other cases, we use the
Gruntz algorithm (see the gruntz() function).
"""
e = sympify(e)
z = sympify(z)
z0 = sympify(z0)
if e == z:
return z0
if e.is_Rational:
return e
if e.is_Pow:
if e.args[0] == z:
if e.args[1].is_Rational:
if e.args[1] > 0:
return z0**e.args[1]
else:
if z0 == 0:
if dir == "+":
return S.Infinity
else:
return -S.Infinity
else:
return z0**e.args[1]
if e.args[1].is_number:
if e.args[1].evalf() > 0:
return S.Zero
else:
if dir == "+":
return S.Infinity
else:
return -S.Infinity
if e.is_Add:
if e.is_polynomial() and z0.is_finite:
return Add(*[limit(term, z, z0, dir) for term in e.args])
else:
# this is a case like limit(x*y+x*z, z, 2) == x*y+2*x
# but we need to make sure, that the general gruntz() algorithm is
# executed for a case like "limit(sqrt(x+1)-sqrt(x),x,oo)==0"
r = e.subs(z, z0)
if r is not S.NaN:
return r
try:
r = gruntz(e, z, z0, dir)
except PoleError:
r = heuristics(e, z, z0, dir)
return r
def limit(e, z, z0, dir="+"):
"""
Compute the limit of e(z) at the point z0.
z0 can be any expression, including oo and -oo.
For dir="+" (default) it calculates the limit from the right
(z->z0+) and for dir="-" the limit from the left (z->z0-). For infinite z0
(oo or -oo), the dir argument doesn't matter.
Examples
========
>>> from sympy import limit, sin, Symbol, oo
>>> from sympy.abc import x
>>> limit(sin(x)/x, x, 0)
1
>>> limit(1/x, x, 0, dir="+")
oo
>>> limit(1/x, x, 0, dir="-")
-oo
>>> limit(1/x, x, oo)
0
Notes
=====
First we try some heuristics for easy and frequent cases like "x", "1/x",
"x**2" and similar, so that it's fast. For all other cases, we use the
Gruntz algorithm (see the gruntz() function).
"""
from sympy import Wild, log
e = sympify(e)
z = sympify(z)
z0 = sympify(z0)
if e == z:
return z0
if e.is_Rational:
return e
if not e.has(z):
return e
# gruntz fails on factorials but works with the gamma function
# If no factorial term is present, e should remain unchanged.
# factorial is defined to be zero for negative inputs (which
# differs from gamma) so only rewrite for positive z0.
if z0.is_positive:
e = e.rewrite(factorial, gamma)
if e.func is tan:
# discontinuity at odd multiples of pi/2; 0 at even
disc = S.Pi / 2
sign = 1
if dir == '-':
sign *= -1
i = limit(sign * e.args[0], z, z0) / disc
if i.is_integer:
if i.is_even:
return S.Zero
elif i.is_odd:
if dir == '+':
return S.NegativeInfinity
else:
return S.Infinity
if e.func is cot:
# discontinuity at multiples of pi; 0 at odd pi/2 multiples
disc = S.Pi
sign = 1
if dir == '-':
sign *= -1
i = limit(sign * e.args[0], z, z0) / disc
if i.is_integer:
if dir == '-':
return S.NegativeInfinity
else:
return S.Infinity
elif (2 * i).is_integer:
return S.Zero
if e.is_Pow:
b, ex = e.args
c = None # records sign of b if b is +/-z or has a bounded value
if b.is_Mul:
c, b = b.as_two_terms()
if c is S.NegativeOne and b == z:
c = '-'
elif b == z:
c = '+'
if ex.is_number:
if c is None:
base = b.subs(z, z0)
if base.is_finite and (ex.is_bounded or base is not S.One):
return base**ex
else:
if z0 == 0 and ex < 0:
if dir != c:
# integer
if ex.is_even:
return S.Infinity
elif ex.is_odd:
return S.NegativeInfinity
# rational
elif ex.is_Rational:
return (S.NegativeOne**ex) * S.Infinity
else:
return S.ComplexInfinity
return S.Infinity
return z0**ex
if e.is_Mul or not z0 and e.is_Pow and b.func is log:
if e.is_Mul:
if abs(z0) is S.Infinity:
n, d = e.as_numer_denom()
# XXX todo: this should probably be stated in the
# negative -- i.e. to exclude expressions that should
# not be handled this way but I'm not sure what that
# condition is; when ok is True it means that the leading
# term approach is going to succeed (hopefully)
ok = lambda w: (z in w.free_symbols and any(
a.is_polynomial(z) or any(
z in m.free_symbols and m.is_polynomial(z)
for m in Mul.make_args(a)) for a in Add.make_args(w)))
if all(ok(w) for w in (n, d)):
u = C.Dummy(positive=(z0 is S.Infinity))
inve = (n / d).subs(z, 1 / u)
return limit(inve.as_leading_term(u), u, S.Zero,
"+" if z0 is S.Infinity else "-")
# weed out the z-independent terms
i, d = e.as_independent(z)
if i is not S.One and i.is_bounded:
return i * limit(d, z, z0, dir)
else:
i, d = S.One, e
if not z0:
# look for log(z)**q or z**p*log(z)**q
p, q = Wild("p"), Wild("q")
r = d.match(z**p * log(z)**q)
if r:
p, q = [r.get(w, w) for w in [p, q]]
if q and q.is_number and p.is_number:
if q > 0:
if p > 0:
return S.Zero
else:
return -oo * i
else:
if p >= 0:
return S.Zero
else:
return -oo * i
if e.is_Add:
if e.is_polynomial() and not z0.is_unbounded:
return Add(*[limit(term, z, z0, dir) for term in e.args])
# this is a case like limit(x*y+x*z, z, 2) == x*y+2*x
# but we need to make sure, that the general gruntz() algorithm is
# executed for a case like "limit(sqrt(x+1)-sqrt(x),x,oo)==0"
unbounded = []
unbounded_result = []
unbounded_const = []
unknown = []
unknown_result = []
finite = []
zero = []
def _sift(term):
if z not in term.free_symbols:
if term.is_unbounded:
unbounded_const.append(term)
else:
finite.append(term)
else:
result = term.subs(z, z0)
bounded = result.is_bounded
if bounded is False or result is S.NaN:
unbounded.append(term)
if result != S.NaN:
# take result from direction given
result = limit(term, z, z0, dir)
unbounded_result.append(result)
elif bounded:
if result:
finite.append(result)
else:
zero.append(term)
else:
unknown.append(term)
unknown_result.append(result)
for term in e.args:
_sift(term)
bad = bool(unknown and unbounded)
if bad or len(unknown) > 1 or len(unbounded) > 1 and not zero:
uu = unknown + unbounded
# we won't be able to resolve this with unbounded
# terms, e.g. Sum(1/k, (k, 1, n)) - log(n) as n -> oo:
# since the Sum is unevaluated it's boundedness is
# unknown and the log(n) is oo so you get Sum - oo
# which is unsatisfactory. BUT...if there are both
# unknown and unbounded terms (condition 'bad') or
# there are multiple terms that are unknown, or
# there are multiple symbolic unbounded terms they may
# respond better if they are made into a rational
# function, so give them a chance to do so before
# reporting failure.
u = Add(*uu)
f = u.normal()
if f != u:
unknown = []
unbounded = []
unbounded_result = []
unknown_result = []
_sift(limit(f, z, z0, dir))
# We came in with a) unknown and unbounded terms or b) had multiple
# unknown terms
# At this point we've done one of 3 things.
# (1) We did nothing with f so we now report the error
# showing the troublesome terms which are now in uu. OR
# (2) We did something with f but the result came back as unknown.
# Normally this wouldn't be a problem,
# but we had either multiple terms that were troublesome (unk and
# unbounded or multiple unknown terms) so if we
# weren't able to resolve the boundedness by now, that indicates a
# problem so we report the error showing the troublesome terms which are
# now in uu.
if unknown:
if bad:
msg = 'unknown and unbounded terms present in %s'
elif unknown:
msg = 'multiple terms with unknown boundedness in %s'
raise NotImplementedError(msg % uu)
# OR
# (3) the troublesome terms have been identified as finite or unbounded
# and we proceed with the non-error code since the lists have been updated.
u = Add(*unknown_result)
if unbounded_result or unbounded_const:
unbounded.extend(zero)
inf_limit = Add(*(unbounded_result + unbounded_const))
if inf_limit is not S.NaN:
return inf_limit + u
if finite:
return Add(*finite) + limit(Add(*unbounded), z, z0, dir) + u
else:
return Add(*finite) + u
if e.is_Order:
args = e.args
return C.Order(limit(args[0], z, z0), *args[1:])
try:
r = gruntz(e, z, z0, dir)
if r is S.NaN:
raise PoleError()
except (PoleError, ValueError):
r = heuristics(e, z, z0, dir)
return r
def limit(e, z, z0, dir="+"):
"""
Compute the limit of e(z) at the point z0.
z0 can be any expression, including oo and -oo.
For dir="+" (default) it calculates the limit from the right
(z->z0+) and for dir="-" the limit from the left (z->z0-). For infinite z0
(oo or -oo), the dir argument doesn't matter.
Examples
========
>>> from sympy import limit, sin, Symbol, oo
>>> from sympy.abc import x
>>> limit(sin(x)/x, x, 0)
1
>>> limit(1/x, x, 0, dir="+")
oo
>>> limit(1/x, x, 0, dir="-")
-oo
>>> limit(1/x, x, oo)
0
Notes
=====
First we try some heuristics for easy and frequent cases like "x", "1/x",
"x**2" and similar, so that it's fast. For all other cases, we use the
Gruntz algorithm (see the gruntz() function).
"""
from sympy import Wild, log
e = sympify(e)
z = sympify(z)
z0 = sympify(z0)
if e == z:
return z0
if e.is_Rational:
return e
if not e.has(z):
return e
# gruntz fails on factorials but works with the gamma function
# If no factorial term is present, e should remain unchanged.
# factorial is defined to be zero for negative inputs (which
# differs from gamma) so only rewrite for positive z0.
if z0.is_positive:
e = e.rewrite(factorial, gamma)
if e.func is tan:
# discontinuity at odd multiples of pi/2; 0 at even
disc = S.Pi / 2
sign = 1
if dir == '-':
sign *= -1
i = limit(sign * e.args[0], z, z0) / disc
if i.is_integer:
if i.is_even:
return S.Zero
elif i.is_odd:
if dir == '+':
return S.NegativeInfinity
else:
return S.Infinity
if e.func is cot:
# discontinuity at multiples of pi; 0 at odd pi/2 multiples
disc = S.Pi
sign = 1
if dir == '-':
sign *= -1
i = limit(sign * e.args[0], z, z0) / disc
if i.is_integer:
if dir == '-':
return S.NegativeInfinity
else:
return S.Infinity
elif (2 * i).is_integer:
return S.Zero
if e.is_Pow:
b, ex = e.args
c = None # records sign of b if b is +/-z or has a bounded value
if b.is_Mul:
c, b = b.as_two_terms()
if c is S.NegativeOne and b == z:
c = '-'
elif b == z:
c = '+'
if ex.is_number:
if c is None:
base = b.subs(z, z0)
if base != 0 and (ex.is_bounded or base is not S.One):
return base**ex
else:
if z0 == 0 and ex < 0:
if dir != c:
# integer
if ex.is_even:
return S.Infinity
elif ex.is_odd:
return S.NegativeInfinity
# rational
elif ex.is_Rational:
return (S.NegativeOne**ex) * S.Infinity
else:
return S.ComplexInfinity
return S.Infinity
return z0**ex
if e.is_Mul or not z0 and e.is_Pow and b.func is log:
if e.is_Mul:
if abs(z0) is S.Infinity:
n, d = e.as_numer_denom()
# XXX todo: this should probably be stated in the
# negative -- i.e. to exclude expressions that should
# not be handled this way but I'm not sure what that
# condition is; when ok is True it means that the leading
# term approach is going to succeed (hopefully)
ok = lambda w: (z in w.free_symbols and any(
a.is_polynomial(z) or any(
z in m.free_symbols and m.is_polynomial(z)
for m in Mul.make_args(a)) for a in Add.make_args(w)))
if all(ok(w) for w in (n, d)):
u = C.Dummy(positive=(z0 is S.Infinity))
inve = (n / d).subs(z, 1 / u)
return limit(inve.as_leading_term(u), u, S.Zero,
"+" if z0 is S.Infinity else "-")
# weed out the z-independent terms
i, d = e.as_independent(z)
if i is not S.One and i.is_bounded:
return i * limit(d, z, z0, dir)
else:
i, d = S.One, e
if not z0:
# look for log(z)**q or z**p*log(z)**q
p, q = Wild("p"), Wild("q")
r = d.match(z**p * log(z)**q)
if r:
p, q = [r.get(w, w) for w in [p, q]]
if q and q.is_number and p.is_number:
if q > 0:
if p > 0:
return S.Zero
else:
return -oo * i
else:
if p >= 0:
return S.Zero
else:
return -oo * i
if e.is_Add:
if e.is_polynomial():
if not z0.is_unbounded:
return Add(*[limit(term, z, z0, dir) for term in e.args])
elif e.is_rational_function(z):
rval = Add(*[limit(term, z, z0, dir) for term in e.args])
if rval != S.NaN:
return rval
if not any([a.is_unbounded for a in e.args]):
e = e.normal() # workaround for issue 3744
if e.is_Order:
args = e.args
return C.Order(limit(args[0], z, z0), *args[1:])
try:
r = gruntz(e, z, z0, dir)
if r is S.NaN:
raise PoleError()
except (PoleError, ValueError):
r = heuristics(e, z, z0, dir)
return r
