question 1 6.25 pts use lhopitals rule to evaluate the limit. lim x→8? -8 -16 8 16 question 2 6.25 pts use…

Explanation:

Step1: Check form for Question 1

Since the limit $\lim_{x\rightarrow8}$ is incomplete in the first - question, we focus on Question 2. For $\lim_{x\rightarrow1}\frac{x^{3}-5x^{2}+4}{x - 1}$, when $x = 1$, the numerator $1^{3}-5\times1^{2}+4=1 - 5 + 4=0$ and the denominator $1-1 = 0$, so it is in the $\frac{0}{0}$ indeterminate form.

Step2: Apply L'Hopital's Rule

L'Hopital's Rule states that if $\lim_{x\rightarrow a}\frac{f(x)}{g(x)}$ is in the $\frac{0}{0}$ or $\frac{\infty}{\infty}$ form, then $\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}$. Differentiate the numerator $f(x)=x^{3}-5x^{2}+4$, $f'(x)=3x^{2}-10x$. Differentiate the denominator $g(x)=x - 1$, $g'(x)=1$. So, $\lim_{x\rightarrow1}\frac{x^{3}-5x^{2}+4}{x - 1}=\lim_{x\rightarrow1}\frac{3x^{2}-10x}{1}$.

Step3: Evaluate the new limit

Substitute $x = 1$ into $\frac{3x^{2}-10x}{1}$. We get $3\times1^{2}-10\times1=3 - 10=-7$.

Answer:

-7