Find the value of $\lim_{x \to 2} \frac {xf^2(x)-9}{(x-2)}$ Suppose that $$\lim_{x\to 2}\frac{x f(x)-3}{x-2}=5.$$
Then, what is the value of $$\lim_{x\to 2}\frac{x f^2(x)-9}{x-2}?$$
 A: You can see that in $\lim_{x\to 2}\frac{x f(x)-3}{x-2}=5$, the denominator goes to zero. Since we have a limit, the numerator should go to zero too and hence
$$\begin{align}
\lim_{x\to 2}x f(x)-3 &= 0 \\
\lim_{x\to 2}x f(x) &= 3 \\
\frac{\lim_{x\to 2}x f(x)}{\lim_{x\to 2}x} &= \frac{3}{\lim_{x\to 2}x} \\
\lim_{x\to 2}\frac{x f(x)}{x} &= \frac{3}{2}
\end{align}$$
Then the limit of $f(x)$ at $x=2$ will be
$$\lim_{x\to 2}f(x)=\frac{3}{2}$$
and then we can conclude that 
$$\begin{align}
\lim_{x\to 2}f(x) \cdot \lim_{x\to 2}f(x) &= \frac{3}{2} \cdot \frac{3}{2} \\
\lim_{x \to 2}f(x) \cdot f(x) &= \frac{9}{4} \\
\lim_{x \to 2}f^2(x) &= \frac{9}{4} \\
\lim_{x \to 2}x \cdot \lim_{x \to 2}f^2(x) &= (\lim_{x \to 2}x) \cdot \frac{9}{4} \\
\lim_{x \to 2} xf^2(x) &= \frac{9}{2} \\
\lim_{x\to 2}xf^2(x)-9 &= -\frac{9}{2}
\end{align}$$
So we can conclude for the second limit that 
$$\lim_{x\to 2^{\mp}}\frac{x f^2(x)-9}{x-2}= \pm \infty$$
as the numerator goes to $-\frac{9}{2}$ and the denominator goes to zero.
A: HINT:
Set $x-2=h$
$$\lim_{h\to0}\dfrac{(h+2)f(h+2)-3}h=5$$
$$\implies(h+2)f(h+2)-3=5h+O(h^2)\iff f(h+2)=\dfrac{3+5h+O(h^2)}{h+2}$$
