Evaluate $\lim_{x \to \infty} \left( \left( \frac{x+1}{x-1} \right)^x - e^2\right) x^2$ $$\underset{x\to \infty}{\lim} \left( \left( \frac{x+1}{x-1} \right)^x - e^2\right) x^2$$
My Attempt:
$$L = \underset{t\to 0}{\lim} \frac{\left( \left( \frac{t+1}{t-1} \right)^{\frac 1t} - e^2\right)} {t^2}$$
I Now have a $\frac 00$ form that I could use L'Hopital rule with, but I don't want to differentiate the ugly looking function in the numerator. Is there an easier way to solve these kinds of problems? Maybe a taylor series expansion for $(1+t)^{\frac 1t}, t \to 0$ forms would come in handy here and I could just subtract the $e^2$ from the resulting expansion.
 A: Let $y=\left( \frac{t+1}{t-1} \right)^{\frac 1t}$. Then,
$$\ln y = \frac1t \ln \frac{t+1}{t-1} = \frac1t [\ln (1+t)-\ln(1-t)]$$
$$=  \frac1t [(t-\frac12t^2+\frac13t^3) - (t+\frac12t^2-\frac13t^3) +O(t^5)] =2+\frac23t^2+O(t^4)$$
and
$$\left( \frac{t+1}{t-1} \right)^{\frac 1t}=y= e^{2+\frac23t^2+O(t^4)} = e^2(1+\frac23t^2+O(t^4))$$
Thus,
$$L = \underset{t\to 0}{\lim} \frac{\left( \frac{t+1}{t-1} \right)^{\frac 1t} - e^2} {t^2} = \underset{t\to 0}{\lim} \frac{e^2(1+\frac23t^2+O(t^4))-e^2}{t^2}
= \underset{t\to 0}{\lim} \left(\frac23e^2+O(t^2)\right)=\frac23e^2$$
A: $$ \left( \left( \frac{x+1}{x-1} \right)^x - e^2\right) x^2$$
Start with
$$y=\left( \frac{x+1}{x-1} \right)^x\implies \log(y)=x\log\left( \frac{x+1}{x-1} \right)=x\log\left(1+ \frac{2}{x-1} \right)$$ Now, by Taylor
$$\log(y)=x\left(\frac{2}{x}+\frac{2}{3 x^3}+\frac{2}{5 x^5}+O\left(\frac{1}{x^7}\right)\right)={2}+\frac{2}{3 x^2}+\frac{2}{5 x^4}+O\left(\frac{1}{x^6}\right)$$
$$y=e^{\log(y)}=e^2+\frac{2 e^2}{3 x^2}+\frac{28 e^2}{45
   x^4}+O\left(\frac{1}{x^6}\right)$$
$$(y-e^2)x^2=\frac{2 e^2}{3}+\frac{28 e^2}{45 x^2}+O\left(\frac{1}{x^4}\right)$$
A: First,you made a mistake.$L$ should be
$$ \lim_{t\to 0}\dfrac{\left(\dfrac{1+t}{1-t}\right)^{\dfrac{1}{t}}-e^2}{t^2}$$
Then,we could use equivalent substitution to compute the limit as follow.
$$ln(1+t)=t-\dfrac{t^2}{2}+\dfrac{t^3}{3}+o(t^4)$$
$$\dfrac{1}{t}(ln(1+t)-ln(1-t))=2+\dfrac{2t^2}{3}+o(t^2)$$
$$\lim_{t\to 0}\dfrac{1}{t}ln\left( \dfrac{1+t}{1-t}\right)-2=0$$
$$L=\lim_{t\to 0}\dfrac{( e^{\dfrac{1}{t}ln\left( \dfrac{1+t}{1-t}\right)-2}-1) e^2}{t^2}=\lim_{t\to 0}(\dfrac{1}{t}ln\left(\dfrac{1+t}{1-t}\right)-2)\frac{e^2}{t^2}$$
$$=\lim_{t\to 0}\frac{e^2}{t^2}(\dfrac{2t^2}{3})=\dfrac{2e^2}{3} $$
