Solve limit $\lim_{x\to1}\frac{x^2-1}{\ln(x)}$ without using L'Hôpital's rule? How to solve this limit?
$$\lim_{x\to1}\frac{x^2-1}{\ln(x)}=\frac00$$
 A: Set $x-1=h\iff x=1+h$
$$\lim_{x\to1}\dfrac{x^2-1}{\ln x}=\dfrac{\lim_{h\to0}\dfrac{(h+1)^2-1}h}{\lim_{h\to0}\dfrac{\ln(1+h)}h}=?$$
A: In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities

$$\bbox[5px,border:2px solid #C0A000]{\frac{x-1}{x}\le \log(x)\le x-1} \tag 1$$

for $x>0$.
Hence, we have from $(1)$
$$\frac{x^2-1}{x-1}\le \frac{x^2-1}{\log(x)}\le \frac{x(x^2-1)}{x-1} \tag 2$$
whence application of the squeeze theorem yields the coveted limit

$$\lim_{x\to 1}\frac{x^2-1}{\log(x)}=2$$

A: With substitution $x=e^t$ we have:
$$\lim_{x\to1}\frac{x^2-1}{\ln x}=\lim_{t\to0}\frac{e^{2t}-1}{t}=\lim_{t\to0}\frac{1+2t+\frac{(2t)^2}{2}-1}{t}=2$$
where
$$e^\theta=1+\theta+\frac{\theta^2}{2}$$
A: Notice that
$$\ln(x)=(x-1)-\frac12(x-1)^2+\mathcal O(x^3)$$
and
$$x^2-1=(x-1)(x+1)$$
So the reciprocal of the limit is
$$\frac1L=\lim_{x\to1}\frac1{x+1}-\frac{(x-1)}{2(x+1)}+\mathcal O\left(\frac{(x-1)^2}{x+1}\right)=\frac12$$
So that
$$L=2$$
A: By a shift of the variable your limit is equivalent to
$$\lim_{t\to0}\frac{t(t+1)}{\ln(1+t)}=\frac{\lim_{t\to0}t+1}{\lim_{t\to0}\dfrac{\ln(1+t)}t}=2.$$
