If $\lim_{x\to \infty} \left(1+\frac{a}{x}+\frac{b}{x^2}\right)^{2x} = e^2$, find $a,b\in \mathbb{R}$ If $\displaystyle\lim_{x\to \infty} \left(1+\frac{a}{x}+\frac{b}{x^2}\right)^{2x} = e^2$, find $a,b\in \mathbb{R}$
I tried this by converting the expression given inside the brackets (denote by $f(x)$) into $e^\left({\displaystyle\lim_{x\to \infty}{2x}{\ln(f(x))}}\right)$ and then evaluating the power with $2$. But I didn't really get the correct solution.
I found a solution here but I couldn't understand how they did. I feel that they were along the same lines as I except that either of us did something wrong. 
 A: Set
$$1+\frac{a}{x}+\frac{b}{{{x}^{2}}}=\left( 1+\frac{\alpha }{x} \right)\left( 1+\frac{\beta }{x} \right)=1+\frac{\alpha +\beta }{x}+\frac{\alpha \beta }{{{x}^{2}}}$$
So
$$a=\alpha +\beta \ \ \text{ and }\ \ b=\alpha \beta ,$$
Now
$$\underset{x\to \infty }{\mathop{\lim }}\,{{\left( 1+\frac{a}{x}+\frac{b}{{{x}^{2}}} \right)}^{2x}}=\underset{x\to \infty }{\mathop{\lim }}\,{{\left( 1+\frac{\alpha }{x} \right)}^{2x}}{{\left( 1+\frac{\beta }{x} \right)}^{2x}}={{e}^{2\alpha }}{{e}^{2\beta }}={{e}^{2\left( \alpha +\beta  \right)}}={{e}^{2a}}$$
Hence ${{e}^{2a}}={{e}^{2}}\Rightarrow a=1$ and the valve of $b$  is arbitrary. 
A: Note that we have
$$\left(1+\frac{a}{x}+\frac{b}{x^2}\right)^{2x}=\left[\left(1+\frac{ax+b}{x^2}\right)^{\frac{x^2}{ax+b}}\right]^\frac{2ax^2+2xb}{x^2}\to e^{2}$$
when $a=1$ for every $b$.
A: You can use the Taylor series of $ln(1+u) = u-\frac{u^2}{2}+\frac{u^3}{3} ...$
Thus, letting $u=\frac{a}{x}+\frac{b}{x^2}$ we have  $ln(1+\frac{a}{x}+\frac{b}{x^2})^{2x} = 2x\left( (\frac{a}{x}+\frac{b}{x^2})-(\frac{a}{x}+\frac{b}{x^2})^2 ...\right) = 2a+\frac{2b-a^2}{x} ...$
All further terms will have even higher powers of $x$ in the denominator, so as $x \rightarrow \infty$, they go to zero. Taking the log of the other side, you get $2a = 2e$, so $a = 1$ and b is free.
A: we know that
$$\ln (1+X)\sim X (X\to 0)$$
thus
$$\ln (1+\frac {a}{x}+\frac{b}{x^2})=$$
$$\ln\Bigl(1+\frac {a}{x}(1+\frac {b}{ax})\Bigr)$$
$$\sim \frac{a}{x} \; (x\to +\infty)$$
the limit is surely $$e^{2a} $$
So $a=1$ and $b $ is arbitrary.
A: With $f(x)=1+\frac{a}{x}+\frac{b}{x^2}$ you have when $ x \to + \infty$:
$$\ln(f(x))= \ln \left(1+\frac{a}{x}+o\left(\frac{1}{x} \right) \right)=\frac{a}{x}+o\left(\frac{1}{x}\right)$$
so:
$$ 2x \ln(f(x))= 2a+o(1)$$
i.e:
$$\lim_{x \to + \infty} 2x \ln(f(x))=2a$$
