find $\lim_{x\to 2}\frac{f(x^2-2)-f(f(x)-3)}{(x-2)^2}$ If the equation of the tangent to the graph of the function $y=f(x)$ at $x=2$ is $4x-y-3=0$ and at this point tangent cuts the graph also,then find $\lim_{x\to 2}\frac{f(x^2-2)-f(f(x)-3)}{(x-2)^2}$

$$\lim_{x\to 2}\frac{f(x^2-2)-f(f(x)-3)}{(x-2)^2}$$
As the denominator is tending to zero,so numerator must br tending to zero.
So I applied L Hospital rule two times to get
$$\lim_{x\to 2}\frac{f'(x^2-2).2x-f'(f(x)-3).f'(x)}{2(x-2)}$$ for the first time and $$\lim_{x\to 2}\frac{4x^2f''(x^2-2)+2f'(x^2-2)-f'(f(x)-3).f''(x)-f''(f(x)-3).(f'(x))^2}{2}$$
As the equation of tangent at the point $x=2$ is $y=4x-3$
So slope of tangent at $x=2$ is $4$ i.e. $f'(2)=4$ and as the tangent cuts the graph at $x=2$,so put $x=2$ in $y=4x-3$ to get $y=5$, so $f(2)=5$ but 
I don't know how to find $f''(2)$.

I am stuck here, please help me. Thanks.
 A: From the given information you know that $f(2)=5$, $f'(2)=4$ and $f''(2)=0$ (“the tangent cuts the graph” should mean that $2$ is an inflection point). So you can write
$$
f(2+h)=5+4h+o(h^2)
$$
For the limit, do the substitution $x=2+h$, so it becomes
$$
\lim_{h\to0}\frac{f(2+4h+h^2)-f(f(2+h)-3)}{h^2}
$$
Now you can write
$$
f(2+4h+h^2)=5+4(4h+h^2)+o((4h+h^2)^2)=5+16h+4h^2+o(h^2)
$$
and
$$
f(2+h)-3=5+4h+o(h^2)-3=2+4h+o(h^2)
$$
so
$$
f(f(2+h)-3)=f(2+4h+o(h^2))=5+16h+o(h^2)
$$
Now the limit becomes
$$
\lim_{h\to0}\frac{(5+16h+4h^2+o(h^2))-(5+16h+o(h^2))}{h^2}=4
$$
A: The numerator is approaching 0 as you can see
$$\lim_{x\to 2}\frac{f(x^2-2)-f(f(x)-3)}{(x-2)^2} = \lim_{x\to 2}\frac{f(2^2-2)-f(f(2)-3)}{(x-2)^2} = \lim_{x\to 2}\frac{f(2)-f(5-3)}{(x-2)^2} = \lim_{x\to 2}\frac{f(2)-f(2)}{(x-2)^2}$$
By applying L'Hôpital's rule you have $$\lim_{x\to 2}\frac{f(x^2-2)-f(f(x)-3)}{(x-2)^2} = \lim_{x\to 2}\frac{f'(x^2-2).2x-f'(f(x)-3).f'(x)}{2(x-2)} = \lim_{x\to 2}\frac{f'(2^2-2) \cdot 2x-f'(f(2)-3) \cdot f'(2)}{2(x-2)} = \lim_{x\to 2}\frac{f'(2) \cdot 2x-f'(5-3) \cdot f'(2)}{2(x-2)} = \lim_{x\to 2}\frac{4 \cdot 2x-4 \cdot 4}{2(x-2)} = 4 \lim_{x\to 2}\frac{x-2}{x-2} = 4$$
A: In general, the equation of a tangent is $y-f(x_0)=f'(x_0)(x-x_0)$.
Rearrange $4x-y-3=0$, we have $y-5=4(x-2)$,
i.e. $f(2)=5$ and $f'(2)=4$.
As $t\rightarrow 2$, $f(x^2-2)-f(f(x)-3) \rightarrow f(4-2)-f(5-3)=0$.
Now 
$\lim_{x\rightarrow 2} \frac{f(x^2-2)-f(f(x)-3)}
                            {(x-2)^2}
=\lim_{x\rightarrow 2} \frac{f'(x^2-2)\cdot 2x-f'(f(x)-3)\cdot f'(x)}
                            {2(x-2)} $
As $t\rightarrow 2$, $f'(x^2-2)\cdot 2x-f'(f(x)-3)\cdot f'(x) \rightarrow f'(2)\times 4-f'(5-3)\times 4=0$.
Then applying  L'Hospital rule once more,
$\begin{eqnarray*}
   &&
   \lim_{x\rightarrow 2}
   \frac{f(x^2-2)-f(f(x)-3)}{(x-2)^2} \\
   &=&
   \lim_{x\rightarrow 2}
   \frac{f''(x^2-2) \cdot (2x)^2+f'(x^2-2)\cdot 2-
         f''(f(x)-3)\cdot [f'(x)]^2-f'(f(x)-3)\cdot f''(x)}{2} \\
   &=&
   \frac{f''(2)(4)^2+f'(2)\times 2-f''(2)[f'(2)]^2-f'(2)\times f''(2)}{2} \\
   &=& 4-2f''(2)
\end{eqnarray*}$
Simply leaves $f''(2)$ in your answer.
