Find $f(0),f'(0),f''(0)$ 
Suppose $f(x)$ has second derivative on some neighbourhood around $x=0$, and $$\lim_{x\to 0}\left(1+x+\frac{f(x)}{x}\right)^{1/x}=e^3$$ Find $f(0),f'(0),f''(0)$

I know $f(0)$ and $f'(0)$ must be $0$ or the limit wouldn't exist. Then after some rough calculation, I guess $f''(0)$ should be $2$. My problem is I have trouble to write the proof with strict words.
 A: By the given condition we get via taking logarithm (and the fact that $\log$ function is continuous) $$\lim_{x\to 0}\frac{1}{x}\log\left(1+x+\frac{f(x)}{x}\right)=3\tag{1}$$ and by multiplying with $x$ we can see that $$\lim_{x\to 0}\log\left(1+x+\frac {f(x)} {x} \right) =0$$ Next the continuity of $\exp$ function gives us $$\lim_{x\to 0}\left(1+x+\frac{f(x)}{x}\right)=1$$ so that $$\lim_{x\to 0}\frac{f(x)}{x}=0\tag{2}$$ Multiplication by $x$ gives us $$f(0)=\lim_{x\to 0}f(x)=\lim_{x\to 0}x\cdot \frac {f(x)} {x} =0\cdot 0=0$$ and then $$f'(0)=\lim_{x\to 0}\frac {f(x) - f(0)}{x}=\lim_{x\to 0}\frac{f(x)}{x}=0\text{ (via (2))}$$ To evaluate $f''(0)$ we need more effort. It is known that the second derivative $f''(0)$ exists and hence the limit $$f''(0)=\lim_{x\to 0}\frac{f'(x)-f'(0)}{x}=\lim_{x\to 0}\frac{f'(x)}{x}$$ exists. By L'Hospital's Rule we have $$f''(0)=2\lim_{x\to 0}\frac{f(x)}{x^2}\tag{3}$$ It is now time to revisit equation $(1)$. Using the fundamental limit $$\lim_{t\to 0}\frac {\log(1+t)} {t} =1$$ the equation $(1)$ can be rewritten as $$\lim_{x\to 0}\dfrac{x+\dfrac{f(x)} {x} }{x}=3$$ or $$\lim_{x\to 0}\frac{f(x)}{x^2}=2$$ and by equation $(3)$ we see that $f''(0)=4$.
To sum up $$f(0)=f'(0)=0,f''(0)=4$$ As the above solution shows we don't really need the existence of second derivative $f''$ in some neighborhood of $0$ but only its existence at the point $0$.
A: First， as you see, $f(0)=f'(0)=0$. By taking logarithm, we rewrite the assumption as follows
$$\lim\limits_{x\to0}\frac{\ln(1+x+\frac{f(x)}{x})}{x}=3.$$ Then, by L'Hôpital's rule， we have
$$\lim\limits_{x\to 0}\frac{1}{1+x+\frac{f(x)}{x}}\left(1+\frac{f'(x)x-f(x)}{x^2}\right)=3.$$ Noting that $\lim\limits_{x\to0}f(x)/x=0$, the above limit implies that
$$\lim\limits_{x\to 0}\frac{f'(x)x-f(x)}{x^2}=2.$$ Using L'Hôpital's rule once more, we can get
$$f''(0)=4.$$
A: We know that
$$\lim_{x\to0}(1+3x)^{1/x}=e^3,$$ so that $f(x)=2x^2$ fulifills the criterion.
We can show that $$\left(1+x+\frac{a+bx+2x^2}x\right)^{1/x}$$ diverges for any $a,b\ne0$.
A: We can use  it  this to solve this problem.
first we can use this limit representation of $e^k$
$$ 
{
e^k=\lim_{n\to\infty} \left( 1+\frac{k}{n} \right)^n 
\qquad (1)
}
$$
 and changing n to 1 / x and substituting k = 3
$$ 
{
e^3=\lim_{x\to0} \left( 1+
3x \right)^{\frac{1}{x}} 
\qquad (2)
}
$$
Then using (2) we can see that
$$\lim_{x\to 0}(1+x+\frac{f(x)}{x})^\frac{1}{x}=\lim_{x\to0} \left( 1+
3x \right)^{\frac{1}{x}} $$
And afther that we see that 
$$x+\frac{f(x)}{x}=3x$$
$$\frac{f(x)}{x}=2x$$
$$f(x)=2x^2$$
Although this form of equating terms can not be used habitually, since the limit of two funcuons are equal does not mean that two functions are the same. But in this case it can be deduced from the statement that all possible functions f have the same derivative in 0, since otherwise the statement would contradict itself and have multiple solutions. Then with analyzing the derivatives of the simplest function that can be f is sufficient to extrapolate it to all the f.
So we can easily solve the problem by replacing  the zeros and  the derivatives
$$f(0)=2\cdot 0^2 =0$$
$$f'(0)=4\cdot 0 =0 $$
$$f''(0)=4$$
Thanks for reading and goodbye.
