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!
 A: Given
$$\lim_{x\rightarrow 0} \left(1+\frac{x}{(x-1)^2}\right)^{\frac{1}{\sqrt{1+x}-1}}$$ we have with $u= e ^{ \ln u}$,
$$ \lim_{x \to 0} \left(\frac{x}{\left(x - 1\right)^{2}} + 1\right)^{\frac{1}{\sqrt{x + 1} - 1}}  =  \lim_{x \to 0} e^{\ln{\left(\left(\frac{x}{\left(x - 1\right)^{2}} + 1\right)^{\frac{1}{\sqrt{x + 1} - 1}} \right)}} =$$
$$= \lim_{x \to 0} e^{\ln{\left(\left(\frac{x}{\left(x - 1\right)^{2}} + 1\right)^{\frac{1}{\sqrt{x + 1} - 1}} \right)}}  = \lim_{x \to 0} e^{\frac{\ln{\left(\frac{x}{\left(x - 1\right)^{2}} + 1 \right)}}{\sqrt{x + 1} - 1}} $$
Since we have an indeterminate form of type $\left(\frac 00\right)$, we can apply the l'Hopital's rule:
$$e^{ \lim_{x \to 0} \frac{\frac{d}{dx}\left(\ln{\left(\frac{x}{\left(x - 1\right)^{2}} + 1 \right)}\right)}{\frac{d}{dx}\left(\sqrt{x + 1} - 1\right)}}  = e^{ \lim_{x \to 0} \frac{2 \sqrt{x + 1} \left(- \frac{2 x}{\left(x - 1\right)^{3}} + \frac{1}{\left(x - 1\right)^{2}}\right)}{\frac{x}{\left(x - 1\right)^{2}} + 1}}= $$
$$=e^{ \lim_{x \to 0} \frac{2 \sqrt{x + 1} \left(- \frac{2 x}{\left(x - 1\right)^{3}} + \frac{1}{\left(x - 1\right)^{2}}\right)}{\frac{x}{\left(x - 1\right)^{2}} + 1}}  = e^{ \lim_{x \to 0}\left(- \frac{2 \left(x + 1\right)^{\frac{3}{2}}}{\left(x - 1\right) \left(x + \left(x - 1\right)^{2}\right)}\right)}= e^{2} $$
A: $$\begin{align*}
\lim_{x\rightarrow 0} \left(1+\frac{x}{(x-1)^2}\right)^{\frac{1}{\sqrt{1+x}-1}} &= \lim_{x\to0} \exp\left(\ln\left(\left(1+\frac x{(x-1)^2}\right)^{\frac1{\sqrt{x+1}-1}}\right)\right) \tag{1} \\[1ex]
&= \exp\left(\lim_{x\to0} \frac{\ln\left(1+\frac x{(x-1)^2}\right)}{\sqrt{x+1}-1}\right) \tag{2} \\[1ex]
&= \exp\left(\lim_{x\rightarrow 0} \frac{\ln\left(1+\frac x{(x-1)^2}\right) \left(\sqrt{1+x}+1\right)}x\right) \tag{3} \\[1ex]
&= \exp\left(2\lim_{x\rightarrow 0} \frac{\ln\left(1+\frac x{(x-1)^2}\right)}x\right) \tag{4} \\[1ex]
&= \exp\left(-2 \lim_{x\to0} \frac{x+1}{(x-1)(x^2-x+1)}\right) \tag{5} \\[1ex]
&= \exp(-2\times-1) = \boxed{e^2}
\end{align*}$$


*

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

*$(2)$ : 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

A: Okay so we wish to compute the limit
$$L=\lim_{x\rightarrow 0} \left(1+\frac{x}{(x-1)^2}\right)^{\frac{1}{\sqrt{1+x}-1}}.$$
Start by making the substitution $u=\frac{x}{(x-1)^2}$. Then $u\to0$ as $x\to0$, and for sufficiently small $u$ this substitution can be inverted as
$$x=1+\frac{1-\sqrt{4u+1}}{2u}.$$
Furthermore then
$$\frac{1}{\sqrt{1+x}-1}=\frac{1}{\sqrt{2+\frac{1-\sqrt{4u+1}}{2u}}-1}=\frac{1}{u}\cdot\frac{u}{\sqrt{2+\frac{1-\sqrt{4u+1}}{2u}}-1}.$$
Now you can (hopefully) check for yourself that
$$\lim_{u\to0}\frac{u}{\sqrt{2+\frac{1-\sqrt{4u+1}}{2u}}-1}=2.$$
Combining this with the well known limit
$$\lim_{u\to0}(1+u)^{\frac{1}{u}}=e$$
we finally get that
$$L=\lim_{u\to0}\left(\left(1+u\right)^{\frac{1}{u}}\right)^{\frac{u}{\sqrt{2+\frac{1-\sqrt{4u+1}}{2u}}-1}}=e^2.$$
