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 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}
$.
