exercises 2.3 the limit laws\nscore: 12/21 answered: 12/21\nquestion 13\nevaluate the limit: $lim_{h \to…

exercises 2.3 the limit laws\nscore: 12/21 answered: 12/21\nquestion 13\nevaluate the limit: $lim_{h \to 0}\frac{(-4 + h)^{2}-16}{h}$\nquestion help: video message instructor

exercises 2.3 the limit laws\nscore: 12/21 answered: 12/21\nquestion 13\nevaluate the limit: $lim_{h \to 0}\frac{(-4 + h)^{2}-16}{h}$\nquestion help: video message instructor

Answer

Explanation:

Step1: Expand the numerator

Expand $(-4 + h)^2-16$ using $(a + b)^2=a^{2}+2ab + b^{2}$. Here $a=-4$ and $b = h$. So $(-4 + h)^2=16-8h+h^{2}$, and $(-4 + h)^2-16=16-8h+h^{2}-16=h^{2}-8h$.

Step2: Simplify the fraction

The original limit $\lim_{h\rightarrow0}\frac{(-4 + h)^2-16}{h}=\lim_{h\rightarrow0}\frac{h^{2}-8h}{h}$. Factor out an $h$ from the numerator: $\frac{h^{2}-8h}{h}=\frac{h(h - 8)}{h}=h - 8$ for $h\neq0$.

Step3: Evaluate the limit

Now, find $\lim_{h\rightarrow0}(h - 8)$. Substitute $h = 0$ into $h-8$. We get $0-8=-8$.

Answer:

$-8$