Question

evaluate the limit: $lim_{h
ightarrow0}\frac{(8 + h)^{2}-64}{h}$

Explanation:

Step1: Expand the numerator

Expand \((8 + h)^2-64\). Using the formula \((a + b)^2=a^{2}+2ab + b^{2}\), where \(a = 8\) and \(b=h\), we get \((8 + h)^2-64=(64 + 16h+h^{2})-64=16h + h^{2}\).

Step2: Simplify the fraction

The original limit \(\lim_{h
ightarrow0}\frac{(8 + h)^2-64}{h}\) becomes \(\lim_{h
ightarrow0}\frac{16h+h^{2}}{h}\). Factor out an \(h\) from the numerator: \(\lim_{h
ightarrow0}\frac{h(16 + h)}{h}\). Cancel out the common - factor \(h\) (since \(h
eq0\) when taking the limit), we get \(\lim_{h
ightarrow0}(16 + h)\).

Step3: Evaluate the limit

Substitute \(h = 0\) into \(16 + h\). We have \(16+0=16\).

Answer:

16