If $\lim_{x\to 0} (2+(f(x)/x^2))^{1/x} = e^2$ and $f(x)=a_0 + a_1x+a_2x^2...$, then prove that $a_2+|a_3|=1$. 
If $$\lim\limits_{x\to 0} \left(2+\frac{f(x)}{x^2}\right)^{\frac 1x} = e^2$$ and $f(x)=a_0 + a_1x+a_2x^2...$, then prove that $$a_2+|a_3|=1\,.$$

The function seems to be of the form $\infty^{\infty}$.
I don’t know how to start solving it, since it an indeterminate form I don’t know how to evaluate. Can I gain some insight into the question?
EDIT:-
$$y=\lim_{x\to 0} (2+\frac{f(x)}{x^2})^{\frac 1x}$$
$$\log y = \lim_{x\to 0} \frac 1x \log (2+\frac{f(x)}{x^2})$$
$$\log y = \frac 1x (1+\frac{f(x)}{x^2})=2$$
 A: This is not a difficult question, but a rigorous answer requires some effort and an understanding of limit laws.
As mentioned in my comment to the question I will assume that $$f(x) =a_0+a_1x+a_2x^2+a_3x^3+o(x^3)\tag{1}$$ as $x\to 0$. For those not well versed with the little-o notation the above equation is by definition equivalent to $$\lim_{x\to 0}\frac{f(x) - a_0-a_1x-a_2x^2-a_3x^3}{x^3}=0\tag{2}$$ The given limit condition in question is equivalent to $$\lim_{x\to 0}\frac{\log(2+(f(x)/x^2))}{x}=2\tag{3}$$ (this uses the fact that log is continuous and invertible).
And then $$\log\left(2+\frac{f(x)}{x^2}\right)=x\cdot\frac{\log(2+(f(x)/x^2))}{x}\to 0\cdot 2=0$$ as $x\to 0$. Exponentiating  this we see that $$2+\frac{f(x)}{x^2}\to 1$$ as $x\to 0$ (this works because exponentiation is also continuous) or $$\frac{f(x)}{x^2}\to - 1\tag{4}$$ as $x\to 0$. Multiplying by $x$ and $x^2$ we see that $f(x) /x\to 0$ and $f(x) \to 0$.
Now using $(1)$ or $(2)$ with $f(x)\to 0$ we get $a_0=0$. Next using $a_0=0,f(x)/x\to 0$ with $(1)$ we get $a_1=0$. And in similar fashion we get $a_2=-1$ using $(1)$ and $(4)$.
To get $a_3$ we need to use equation $(3)$ also. Since the limit in $(3)$ is non-zero it follows from $(4)$ that $t=1+(f(x)/x^2)\to 0, t\neq 0$ as $x\to 0$ and $(3)$ can then be rewritten as $$\lim_{x\to 0}\frac{t}{x}\cdot\frac{\log(1+t)}{t}=2$$ Since $(\log(1+t))/t\to 1$ as $t\to 0$ it follows that $$\frac{t} {x} =\frac{1}{x}+\frac{f(x)}{x^3}\to 2\tag{5}$$ Using $(2)$ and $(5)$ along with the values $$a_0=a_1=0,a_2=-1$$ we get $a_3=2$.

The flaw in your approach is that you remove the limit operator without any justification. This simply does not work and can not be guaranteed.
A: Lemma: If $\lim_\limits{x\rightarrow 0}(1+g(x))^{\frac{1}{x}}=e^2$, then $\lim_\limits{x\rightarrow 0}g(x)=0$.
Proof of the Lemma: First, we know that $1+g(x)>0$ (see:https://www.wolframalpha.com/input/?i=%281%2Bx%29%5E%7B1%2Fx%7D)
If $\lim_\limits{x\rightarrow 0}g(x)\neq 0$, by definition of limit, there exists $\epsilon_0>0$ and sequence $x_n\rightarrow 0$ and a small enough $\delta$ (say $\delta=10^{-10}$, anyway) when $|x_n|<\delta$, we have $g(x_n)\geq \epsilon_0$ or $g(x_n)\leq -\epsilon_0$.
Now we consider the part of $\delta>x_n>0$ to construct a contradiction.
(i) if $g(x_n)\geq \epsilon_0$, then
$$(1+g(x_n))^{\frac{1}{x_n}}\geq(1+\epsilon_0)^{\frac{1}{x_n}}\geq(1+\epsilon_0)^{\delta^{-1}}>e^2$$ ($t^{1/x_n}$ is increasing, $\delta$ small enough)
(ii) if $g(x_n)\leq -\epsilon_0$
$$0<(1+g(x_n))^{\frac{1}{x_n}}\leq(1-\epsilon_0)^{\frac{1}{x_n}}\leq(1-\epsilon_0)^{\delta^{-1}}<e^2$$
($t^{1/x_n}$ is increasing, $\delta$ small enough)
Contradiction. The lemma is proved.
Now back to the problem. By the lemma we have $\frac{f(x)}{x^2}+1$ goes to $0$, that is
$$\frac{a_0}{x^2}+\frac{a_1}{x}+a_2+a_3x+o(x)+1\rightarrow 0$$
So $a_0=a_1=0$, $a_2=-1$.
Now
$$(2+\frac{f(x)}{x^2})^{1/x}=(1+a_3x+o(x))^{1/x}=\\
(1+a_3 x+o(x))^{\frac{1}{a_3 
 x+o(x)}\cdot~~~~~\frac{a_3 
 x+o(x)}{x}}\rightarrow e^{a_3}=e^2$$
So $a_3=2$
