Finding the limit of function - irrational function How can I find the following limit: 
$$ \lim_{x \rightarrow -1 }\left(\frac{1+\sqrt[5]{x}}{1+\sqrt[7]{x}}\right)$$

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.
