Find $\lim_{x \to 1} \frac {x-1}{\log_e x} $

Question

:$$ L= \lim_{x \to 1} \frac {x-1}{\log_e x} $$

Let $ x = h + 1, h = x - 1. $ as $ x \to 1, h \to 0$

$$L = \lim_{h \to 0} \frac{h} {\log_e (h+1)}$$

here we have a formula $$ \lim_{x \to 0} \frac{\log(1+x)}{x} = 1 $$ can i use it here!?

Answer 1

Its just $$\frac{1}{\frac{log_e(1+h)}{h}}=\frac{1}{1}=1$$

Answer 2

Yes, you can use $\lim_{x\to 0}\frac{\log(1+x)}{x}=1$ by taking inverse as follows $$\lim_{h\to 0}\frac{h}{\log(h+1)}=\lim_{h\to 0}\left(\frac{\log(1+h)}{h}\right)^{-1}=\left(\lim_{h\to 0}\frac{\log(1+h)}{h}\right)^{-1}=(1)^{-1}=\color{red}{1}$$

Answer 3

I thought it might be instructive to present another way forward.

In THIS ANSWER and THIS ONE I showed, without the use of calculus, that the logarithm function satisfies the inequalities

$$\frac{x-1}{x}\le \log(x)\le x-1$$

for $x>0$. Therefore, we can write for $x>1$

$$1 \le \frac{x-1}{\log(x)}\le x$$

and for $0<x<1$

$$x \le \frac{x-1}{\log(x)}\le 1$$

whereupon applying the squeeze theorem yields the result $1$.