Consider the given limit \(\lim_{x->2}\frac{5\cdot x^2-20}{x-2}\)

Note that, \(f'(a)=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}\)

That implies, \(f(x)=5x^2\) and \(a=2\)

Now obtain the limit as follows

\(\lim_{x->2}\frac{5\cdot x^2-20}{x-2}=\lim_{x->2}\frac{5(x^2-4)}{x-2}\)

\(=\lim_{x\rightarrow2}\frac{5(x^2-2^2)}{x-2}\)

\(=\lim_{x\rightarrow2}\frac{5(x-2)(x+2)}{(x-2)}\)

\(=\lim_{x\rightarrow2}5(x+2)\)

\(=5(2+2)\)

\(=5\cdot4\)

\(=20\)

Hence, the value of the given limit is 20.

Note that, \(f'(a)=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}\)

That implies, \(f(x)=5x^2\) and \(a=2\)

Now obtain the limit as follows

\(\lim_{x->2}\frac{5\cdot x^2-20}{x-2}=\lim_{x->2}\frac{5(x^2-4)}{x-2}\)

\(=\lim_{x\rightarrow2}\frac{5(x^2-2^2)}{x-2}\)

\(=\lim_{x\rightarrow2}\frac{5(x-2)(x+2)}{(x-2)}\)

\(=\lim_{x\rightarrow2}5(x+2)\)

\(=5(2+2)\)

\(=5\cdot4\)

\(=20\)

Hence, the value of the given limit is 20.