Finding the limit of function - irrational function How can I find the following limit: 
$$ \lim_{x \rightarrow -1 }\left(\frac{1+\sqrt[5]{x}}{1+\sqrt[7]{x}}\right)$$
 A: Let $x=t^{35}$. As $x \to -1$, we have $t \to-1$. Hence,
$$\lim_{x \to -1} \dfrac{1+\sqrt[5]{x}}{1+\sqrt[7]{x}} = \lim_{t \to -1} \dfrac{1+t^7}{1+t^5} = \lim_{t \to -1} \dfrac{(1+t)(1-t+t^2-t^3+t^4-t^5+t^6)}{(1+t)(1-t+t^2-t^3+t^4)}$$
I am sure you can take it from here.
A: Alternate solution: since numerator and denominator each approach 0, we may use L'Hopital's rule.  This gives $\lim_{x\rightarrow -1} \frac{\frac 15 x^{-4/5}}{\frac 17 x^{-6/7}}=\lim_{x\rightarrow -1} \frac{7}{5} x^{6/7-4/5}=\lim_{x\rightarrow -1} \frac{7}{5}x^{2/35}=\frac{7}{5}$
A: A useful principle when trying to find a limit is
"always expand around zero".
Since we want to see what happens
when $x \to -1$,
let $x = y-1$ so that
we can look at what happens when $y \to 0$.
The expression becomes
$\frac{1+\sqrt[5]{y-1}}{1+\sqrt[7]{y-1}}$.
For an odd value of $n$,
since $y-1 < 0$ (for small $y$),
$\sqrt[n]{y-1} 
= - \sqrt[n]{1-y}
$.
By the binomial theorem,
as $y \to 0$,
$\sqrt[n]{1-y}
\approx 1-y/n$.
Putting this in for $n = 5$ and $n = 7$,
$\frac{1-\sqrt[5]{1-y}}{1-\sqrt[7]{1-y}}
\approx \frac{1-(1-y/5)}{1-(1-y/7)}
= \frac{y/5}{y/7}
= \frac{7}{5}
$.
A: L'Hopital will work on this also, since it is an indeterminate ratio of the form $0/0$.  We get
$$\lim_{x \rightarrow -1} \frac{(1 + x^{1/5})'}{(1 + x^{1/7})'} , $$
which ceases to be indeterminate after simplification...
