Find the limit of function as $x$ approaches $1$ Find $$\lim_{x \to1}\frac{x^{1/13}-x^{1/7}}{x^{1/5}-x^{1/3}}.$$ I tried taking some powers common but it did not help.
 A: Since for $\alpha > 0$ we have:
$$\lim_{x\to 1}\frac{x^{\alpha}-1}{x-1}=\alpha, $$
it follows that:
$$\lim_{x\to 1}\frac{x^{1/13}-x^{1/7}}{x^{1/5}-x^{1/3}}=\lim_{x\to 1}\frac{\frac{x^{1/13}-1}{x-1}-\frac{x^{1/7}-1}{x-1}}{\frac{x^{1/5}-1}{x-1}-\frac{x^{1/3}-1}{x-1}}=\frac{\frac{1}{13}-\frac{1}{7}}{\frac{1}{5}-\frac{1}{3}}= \color{red}{\frac{45}{91}}.$$
A: This would be so much easier with L'Hopital.
$$\frac{\frac{x^{1/13}-1}{x-1}-\frac{x^{1/7}-1}{x-1}}{\frac{x^{1/5}-1}{x-1}-\frac{x^{1/3}-1}{x-1}}$$
A: take $x$ as $(1+h)$ where $h$ tends to zero , therefore $x^{\frac{1}{13}}\approx 1+\frac{h}{13}$ . you get this by binomial expansion where you can neglect the higher power terms as they do not affect the limit. therefore answer is $$\frac{(1+\frac{h}{13})-(1+\frac{h}{7})}{(1+\frac{h}{5})-(1+\frac{h}{3})}=\frac{45}{91}$$
A: $$\lim_{x \to1}\frac{x^{1/13}-x^{1/7}}{x^{1/5}-x^{1/3}}$$
$$\lim_{x \to1}\frac{x^{1/13}\left(x^{1/7-1/13}-1\right)}{x^{1/5}\left(x^{1/3-1/5}-1\right)}$$
$$=\lim_{x\to1}x^{1/13-1/5}\cdot\lim_{x\to1}\frac{x^{6/91}-1}{x^{2/15}-1}$$
$$=1^{1/13-1/5}\cdot\lim_{x\to1}\frac{x^{6/91}-1}{x^{2/15}-1}$$
Now as lcm$(91,15)=91\cdot15,$ set  $x^{1/91\cdot15}=y$
$$\implies\lim_{x\to1}\frac{x^{6/91}-1}{x^{2/15}-1}=\lim_{y\to1}\frac{y^{6\cdot15}-1}{y^{2\cdot91}-1}=\lim_{y\to1}\frac{(y-1)(y^{89}+y^{88}+\cdots+y+1)}{(y-1)(y^{181}+y^{180}+\cdots+y+1)}$$
$$=\frac{90}{182}$$ as $y\to1,y\ne1\iff y-1\ne0$
