evaluate the limit: $lim_{x\rightarrow4}\frac{-2x + 8}{x^{2}-16}$
evaluate the limit: $lim_{x\rightarrow4}\frac{-2x + 8}{x^{2}-16}$
Answer
Explanation:
Step1: Factor the expressions
Factor (x^{2}-16) as ((x + 4)(x - 4)) and (-2x+8) as (-2(x - 4)). So the limit (\lim_{x\rightarrow4}\frac{x^{2}-16}{-2x + 8}=\lim_{x\rightarrow4}\frac{(x + 4)(x - 4)}{-2(x - 4)}).
Step2: Cancel out the common factor
Cancel out the common - factor ((x - 4)) (since (x\neq4) when taking the limit), we get (\lim_{x\rightarrow4}\frac{x + 4}{-2}).
Step3: Substitute (x = 4)
Substitute (x = 4) into (\frac{x + 4}{-2}), we have (\frac{4+4}{-2}=\frac{8}{-2}=-4).
Answer:
(-4)