Q:

Limit of the form $0*\infty$

Consider the limit L=$\lim_{x\to\infty}$ $x^2*((\frac{x+1}{x-1})^x-e^2)$.
As $x$ approaches $\infty$, $x^2$ approaches $\infty$ and $(\dfrac{x+1}{x-1})^x$ approaches $e^2$, So we have a $0*\infty$ situation.  I tried resolving this with L'hopital, by writing $x^2$ as $1/(1/x^2))$, But it wasnt quite clear with how to proceed further.  Another Idea I had was to write out the expansion of $(\dfrac{x+1}{x-1})^x$ by writing it as ($1+\dfrac{2x}{x-1}$)^x, and using the binomial theroem, But that didnt help either.  
This was from a high school maths exam and the syllabus DID-NOT include maclaurin/taylor series, so there has to be an easier way.  The answer given is $2/3*e^2$

A:

It appears that you have a typo and you need to replace $1-x$ in denominator with $x-1$.
Assuming that this is so we can see that the expression under limit is of the form $x^2(A-B) $ where both $A, B$ tend to same number $e^2$. We can write the expression as $$x^2B\cdot\frac{(A/B) - 1}{\log(A/B)}\cdot\log(A/B)$$ and the fraction in middle tends to $1$ so that the desired limit equals the limit of $$e^2x^2(\log A-\log B) =e^2x^2\left(x\log\frac{x+1}{x-1}-2\right)$$ Putting $x=1/h$ we can see that $h\to 0^{+}$ and the above expression becomes $$e^2\cdot\frac{\log(1+h)-\log(1-h)-2h}{h^3}$$ You can use Taylor series to show the above tends to $2e^2/3$.
I don't think that there is an approach which avoids L'Hospital's Rule, Taylor series or some equivant approach. 
If you wish to use integrals you can see that $$\log(1+h)-\log(1-h)-2h=\int_{0}^{h}\left(\frac{2}{1-t^2}-2\right)\,dt$$ which can be further rewritten as $$\int_{0}^{h}\frac{2t^2}{1-t^2}\,dt$$ Putting $t=u^{1/3}$ we can see that the above integral reduces to $$\frac{2}{3}\int_{0}^{h^3}\frac{du}{1-u^{2/3}}$$ and by fundamental theorem of calculus the above divided by $h^3$ tends to $2/3$ and we get the answer as $2e^2/3$.

