<s>[INST] I can't find the solution of $\lim_{x\rightarrow 0} \left(1+\frac{x}{(x-1)^2}\right)^{\frac{1}{\sqrt{1+x}-1}}$ I can't find the solution of
$$\lim_{x\rightarrow 0} \left(1+\frac{x}{(x-1)^2}\right)^{\frac{1}{\sqrt{1+x}-1}}$$
Computing for $x$ goes to $0$ it gives a $1^\infty$ type of indeterminate form. I tried to solve it by making it similar to $0/0$ type of indeterminate form by taking log of both sides and writing as
$$\frac{\ln\left(1+\frac{x}{(x-1)^2}\right)}{\sqrt{1+x}-1}$$
Then I applied L'Hôpital's rule to find the limit and find it infinity for $0^+$ and $0$ for $0^-$ but my solution did not match with the answer.
Thanks for help!
 [/INST] $$\begin{align*}
\lim_{x\rightarrow 0} \left(1+\frac{x}{(x-1)^2}\right)^{\frac{1}{\sqrt{1+x}-1}} &= \lim_{x\to0} \exp\left(\ln\left(\left(1+\frac x{(x-1)^2}\right)^{\frac1{\sqrt{x+1}-1}}\right)\right) \tag{1} \\[1ex]
&= \exp\left(\lim_{x\to0} \frac{\ln\left(1+\frac x{(x-1)^2}\right)}{\sqrt{x+1}-1}\right) \tag{2} \\[1ex]
&= \exp\left(\lim_{x\rightarrow 0} \frac{\ln\left(1+\frac x{(x-1)^2}\right) \left(\sqrt{1+x}+1\right)}x\right) \tag{3} \\[1ex]
&= \exp\left(2\lim_{x\rightarrow 0} \frac{\ln\left(1+\frac x{(x-1)^2}\right)}x\right) \tag{4} \\[1ex]
&= \exp\left(-2 \lim_{x\to0} \frac{x+1}{(x-1)(x^2-x+1)}\right) \tag{5} \\[1ex]
&= \exp(-2\times-1) = \boxed{e^2}
\end{align*}$$


*

*$(1)$ : $x = e^{\ln(x)}$

*$(2)$ : continuity of $e^x$; $\ln(a^b)=b\ln(a)$

*$(3)$ : introduce the conjugate $\sqrt{1+x}+1$

*$(4)$ : $\lim\limits_{x\to c} f(x)g(x) = \lim\limits_{x\to c} f(x) \lim\limits_{x\to c}g(x)$

*$(5)$ : L'Hôpital's rule

</s>