evaluate the limit: $lim_{h\rightarrow0}\frac{(8 + h)^{2}-64}{h}$
evaluate the limit: $lim_{h\rightarrow0}\frac{(8 + h)^{2}-64}{h}$
Answer
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\rightarrow0}\frac{(8 + h)^2-64}{h}) becomes (\lim_{h\rightarrow0}\frac{16h+h^{2}}{h}). Factor out an (h) from the numerator: (\lim_{h\rightarrow0}\frac{h(16 + h)}{h}). Cancel out the common - factor (h) (since (h\neq0) when taking the limit), we get (\lim_{h\rightarrow0}(16 + h)).
Step3: Evaluate the limit
Substitute (h = 0) into (16 + h). We have (16+0=16).
Answer:
16