How to find the following limit involving the given integration 
I have tried the problem and got $\lim_{x\to0}4\frac{f(x)}{x}=4$. From this how to proceed?
Since $\displaystyle\lim_{x\to0}4\frac{f(x)}{x}=4$, so $\displaystyle\lim_{x\to0}\frac{1}{4}(4\frac{f(x)}{x})=\frac{1}{4}\lim_{x\to0}4\frac{f(x)}{x}=1\implies\lim_{x\to0}\frac{f(x)}{x}=1$. Is it correct?
 A: Hint: $\lim_{x\to 0} \frac{f(x)}{x}=L$ if and only if $\lim_{x\to 0} \frac{f(kx)}{kx}=L$ (here $k$ is any nonzero constant). This does not quite match the expressions in the first line, so you’ll have to multiply/divide by some constants to get the expression inside $f(\ )$ to match the denominator. Doing this you will find exactly $\lim_{x\to 0} 4 \frac{f(x)}{x}=4$, as you found. Then all you have to do is divide both sides by $4$.
A: Let $g(x) = f(x)-x$. Then $g(x) \to 0$ as $x \to 0$, and $12\frac{g(4x)}{4x}-10\frac{g(2x)}{2x}+2\frac{g(x)}{x} \to 0$ as $x \to 0$. I'll show $\frac{g(x)}{x} \to 0$ as $x \to 0$ (which gives $f(x)/x \to 1$). 
I'll show $\frac{g(x)}{x} \to 0$ as $x \to 0^+$. The following argument (easily modified) works for $x \to 0^-$. For $n \ge 0$, let $\epsilon_n = \sup_{x \in [0,2^{-n}]} |6\frac{g(4x)}{4x}-5\frac{g(2x)}{2x}+\frac{g(x)}{x}|$. We know $\epsilon_n \to 0$ as $n \to \infty$. Take any $x \in (\frac{1}{2},1]$, and let $x_0,x_1,x_2,\dots = \frac{g(x)}{x},\frac{g(x/2)}{x/2},\frac{g(x/4)}{x/4},\dots$. Then, for each $n \ge 0$, $x_{n+2} = 5x_{n+1}-6x_n+\epsilon_n'$ for some $\epsilon_n' \in [-\epsilon_n,\epsilon_n]$. Fix $\epsilon > 0$. Take $N$ large so that $|\epsilon_n'| \le \epsilon$ for $n \ge N$. Then, by the Lemma below, we must have $|x_n| \le 5\epsilon$ for $n \ge N$, for otherwise $0 < \lim_{n \to \infty} 2^{-n}x_n \le \limsup_{x \to 0} g(x)$, a contradiction. We have shown $x_n \to 0$ uniformly in $x$, giving the result.
Lemma: For any $\epsilon > 0$, $(\epsilon_n)_{n \ge 3} \in [-\epsilon,\epsilon]^{n \ge 3}$, and $x_1,x_2 \in \mathbb{R}$, if either $|x_1| > 5\epsilon$ or $|x_2| > 5\epsilon$, then the sequence $(x_n)_{n \ge 3}$ defined by $x_n = 5x_{n-1}-6x_{n-2}+\epsilon_n$ satisfies $\limsup_{n \to \infty} 2^{-n}x_n > 0$.
I'm too lazy to prove the Lemma (seems like annoying casework), but the intuition as to why it's true is as follows. If $\epsilon_n = 0$ for each $n \ge 3$, then $x_n = c2^n+d3^n$ for some $c,d \in \mathbb{R}, (c,d) \not = (0,0)$ (if $c=0,d=0$, then $x_1,x_2 = 0$). And when the $\epsilon_n$'s can be nonzero but still tiny, they'll affect the individual values of the sequence but not its overall growth rate.
