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.

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
$$
\lim_{h\to0}\frac{(5+16h+4h^2+o(h^2))-(5+16h+o(h^2))}{h^2}=4
$$
