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.

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 $$\lim_{x\to 2}\frac{f(x^2-2)-f(f(x)-3)}{(x-2)^2} = \lim_{x\to 2}\frac{f'(x^2-2).2x-f'(f(x)-3).f'(x)}{2(x-2)} = \lim_{x\to 2}\frac{f'(2^2-2) \cdot 2x-f'(f(2)-3) \cdot f'(2)}{2(x-2)} = \lim_{x\to 2}\frac{f'(2) \cdot 2x-f'(5-3) \cdot f'(2)}{2(x-2)} = \lim_{x\to 2}\frac{4 \cdot 2x-4 \cdot 4}{2(x-2)} = 4 \lim_{x\to 2}\frac{x-2}{x-2} = 4$$
