Evaluate $ \displaystyle\lim_{x\to1} \frac{x^{\frac{1}{13}}-x^{\frac{1}{7}}}{x^{\frac{1}{5}}-x^{\frac{1}{3}}} $

Question

Evaluate: $$ \lim_{x\to1} \frac{x^{\frac{1}{13}}-x^{\frac{1}{7}}}{x^{\frac{1}{5}}-x^{\frac{1}{3}}} $$

I tried applying L-hospital, but it keeps on expanding.

I also tried putting $1-x=t$ and then applying binomial expansion as $$ \lim_{t \to 0} (1-t)^n=1-nt $$, but I am not sure if this expansion is valid for fractional values of $n$.

Any hints on how to proceed?

Answer 1

Hint: Let $y=x^{1\over 13\cdot 7\cdot 5\cdot 3}$. Then you have to calculate $$ \lim_{y\to 1}{y^{105} - y^{195}\over y^{273}- y^{455}} = \lim_{y\to 1}{1 - y^{90}\over y^{168}- y^{350}}$$

$$= \lim_{y\to 1}{1 - y^{90}\over y^{168}(1- y^{182})}={90\over 182}$$

Answer 2

You can apply l'Hopital once and obtain the result indeed recall that

$$(x^a)'=ax^{a-1}\to a$$

As an alternative let $x=1+y$ with $y\to 0$ then use binomial first order approximation as $y\to 0$

$$(1+y)^a= 1+ay+o(y)$$

that is

$$\lim_{x\to1} \frac{x^{\frac{1}{13}}-x^{\frac{1}{7}}}{x^{\frac{1}{5}}-x^{\frac{1}{3}}}=\lim_{y\to0} \frac{(1+y)^{\frac{1}{13}}-(1+y)^{\frac{1}{7}}}{(1+y)^{\frac{1}{5}}-(1+y)^{\frac{1}{3}}}=\lim_{y\to0} \frac{\frac{1}{13}y-\frac{1}{7}y+o(y)}{\frac{1}{5}y-\frac{1}{3}y+o(y)}=$$$$=\lim_{y\to0} \frac{\frac{1}{13}-\frac{1}{7}+o(1)}{\frac{1}{5}-\frac{1}{3}+o(1)}= \frac{\frac{1}{13}-\frac{1}{7}}{\frac{1}{5}-\frac{1}{3}}=\frac{\frac{7-13}{91} }{\frac{3-5}{15}}=\frac{45}{91}$$

Answer 3

If you use L'hopital rule, you need to evaluate $\lim\limits_{x\rightarrow 1}\frac{\frac{1}{13}x^{\frac{-12}{13}}-\frac{1}{7}x^{\frac{-6}{7}}}{\frac{1}{5}x^{\frac{-4}{5}}-\frac{1}{3}x^{\frac{-2}{3}}}=\frac{\frac{1}{13}-\frac{1}{7}}{\frac{1}{5}-\frac{1}{3}}=\frac{45}{91}$.