sympy.limit yields incorrect result

[cstoudt]

Python 2.7.9; SymPy 0.7.6

Consider the function g=(6**(n+1)+n+1)/(6**n+n) where x is real and n is an integer

It is easy to show that the limit of g as n->oo is 6

Yet, look at the output of the following code:

from sympy import symbols, limit, oo, Pow
n=symbols('n',integer=True)
x=symbols('x',real=True)
g=(Pow(6,n+1)+n+1)/(Pow(6,n)+n)
limit(g,n,oo)

out[4]: 1

[smichr]

Interestingly, if you make n positive

>>> n=symbols('n',integer=True,positive=True)
>>> limit(g,n,oo)
(6**(n + 1) + n + 1)/(6**n + n)

[raoulb]

Consider the function `g=(6**(n+1)+n+1)/(6**n+n)` where x is real and n is an integer

Some heuristics seem to fail here. Use the superior Gruntz algorithm: ``` In [9]: g Out[9]: (6**(n + 1) + n + 1)/(6**n + n) In [10]: gruntz(g, n, oo) Out[10]: 6 ```

[cstoudt]

According to the documentation, limit uses some heuristics to solve the "easy" cases for speed, and uses the Gruntz algorithm for all other cases. So something is failing in the heuristics.

Interestingly, gruntz also fails for the case that smichr pointed out:

>>> from sympy.series import gruntz
>>> n=symbols('n',integer=True,positive=True)
>>> gruntz(g, n, oo)
(6**(n + 1) + n + 1)/(6**n + n)

[zanzibar7]

Bug traces back to atleast 0.7.2. Also in a reduced version...

In [3]: limit( (6**(n+1)+1)/(6**n+1), n, oo)
Out[3]: 1

Looks like the heuristic is equating n+1 and n when n is large.

[leosartaj]

On some debugging,
This is due to incorrect result by inve.as_leading_term(). It returns 1 rather than 6.
This is because in general

>>> (6**(n+1))._eval_leading_term(n)
1 # incorrect
>>>(6*6**n)._eval_leading_term(n)
6 # correct

This ultimately boils down to _eval_leading_term method in Pow.
I am working on this and will send a pull request shortly.