Why can't I use L'Hospital to evaluate $\lim\limits_{x\to 2^+}\frac{\ln(x^2-3)}{\sinh(x-2)}$?

Question

$$\lim_{x\to 2^+}\frac{\ln(x^2-3)}{\sinh(x-2)}$$

I am confused about how to solve it, because I got two different answers. I first expanded it with MacLaurin expansion and computed the limit, getting $+\infty$. But I also did L'Hospital, getting $4$.

I know by the graph that the answer is $+\infty$, but why I can't use L'Hospital in this case?

Answer 1

A way to do it is observing that $\sinh(x)\sim_0 x$ and $\ln(1+x)\sim_0 x$, thus

$$\lim_{x\to 2^+}\frac{\ln(x^2-4+1)}{\sinh(x-2)}=\lim_{x\to 2^+}\frac{x^2-4}{x-2}=\lim_{x\to 2^+}\frac{(x+2)(x-2)}{x-2}=4$$

Answer 2

Make you life easier using $x=2+y$ making $$\frac{\log(x^2-3)}{\sinh(x-2)}=\frac{\log(1+4y+y^2)}{\sinh(y)}$$ Now, use l'Hospital rule or Taylor series to get the limit of $4$ when $y\to 0$.

Answer 3

sure you can use it. But note that $(\ln(x^2-3)'= 2x/(x^2-3)$ $$\lim_{x\to 2^+}\frac{\ln(x^2-3)}{\sinh(x-2)} =\lim_{x\to 2^+}\frac{2x}{(x^2-3)\cosh(x-2)} =4$$