Finding a limit involving roots without derivatives

Question

I need to find the following limit $$ \lim_{x\to-1}\frac{1+x^{1/7}}{1+x^{1/5}} $$ using no derivatives. I've tried attempting to rationalize or divide by certain polynomials, but nothing has worked. It's simple using L'Hopital's rule, but that involves derivatives.

(I'm also wondering if the technique used to evaluate that limit could be extended to $$ \lim_{x\to-1}\frac{1+x^{1/m}}{1+x^{1/n}} = \frac{n}{m} \qquad m,n\text{ odd} $$ but that's not the main question.)

Answer 1

Let $t=\sqrt[35]{x}$, then $$ \lim_{x\to-1}\frac{1+x^{1/7}}{1+x^{1/5}} = \lim_{t\to-1}\frac{1+t^5}{1+t^7}= \lim_{t\to-1}\frac{(t+1)(t^4-t^3+t^2-t+1)}{(t+1)(t^6-t^5+...-t+1)}= {5\over 7} $$

Answer 2

The key here is the standard limit formula $$\lim_{x\to a} \frac{x^n-a^n} {x-a} =na^{n-1}$$ We have $$\lim_{x\to - 1}\frac{1+x^{1/7}}{1+x^{1/5}}=\lim_{x\to - 1}\dfrac{\dfrac{x^{1/7}-(-1)^{1/7}}{x-(-1)}}{\dfrac{x^{1/5}-(-1)^{1/5}}{x-(-1)}}=\frac{1/7}{1/5}=\frac{5}{7}$$