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

Explanation:

Step1: Recall L'Hopital's Rule

L'Hopital's Rule states that if $\lim_{x\rightarrow a}\frac{f(x)}{g(x)}$ is in the indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}$.

Step2: Check the form for Question 1

For $\lim_{x\rightarrow 8}\frac{x - 8}{x - 8}$, when $x = 8$, it is in the $\frac{0}{0}$ form. Let $f(x)=x - 8$ and $g(x)=x - 8$. Then $f'(x)=1$ and $g'(x)=1$. So $\lim_{x\rightarrow 8}\frac{x - 8}{x - 8}=\lim_{x\rightarrow 8}\frac{1}{1}=1$. But this is wrong, we assume the original - function is not written correctly. Let's assume it's $\lim_{x\rightarrow 8}\frac{\sqrt{x}- \sqrt{8}}{x - 8}$. It is in the $\frac{0}{0}$ form. Differentiate $f(x)=\sqrt{x}-\sqrt{8}=x^{\frac{1}{2}}-\sqrt{8}$ and $g(x)=x - 8$. $f'(x)=\frac{1}{2\sqrt{x}}$ and $g'(x)=1$. Then $\lim_{x\rightarrow 8}\frac{\sqrt{x}-\sqrt{8}}{x - 8}=\lim_{x\rightarrow 8}\frac{\frac{1}{2\sqrt{x}}}{1}=\frac{1}{2\sqrt{8}}=\frac{1}{4\sqrt{2}}$. If we assume the correct - function is $\lim_{x\rightarrow 8}\frac{x - 8}{\sqrt{x}-\sqrt{8}}$. It is in the $\frac{0}{0}$ form. Differentiate $f(x)=x - 8$ and $g(x)=\sqrt{x}-\sqrt{8}=x^{\frac{1}{2}}-\sqrt{8}$. $f'(x)=1$ and $g'(x)=\frac{1}{2\sqrt{x}}$. Then $\lim_{x\rightarrow 8}\frac{x - 8}{\sqrt{x}-\sqrt{8}}=\lim_{x\rightarrow 8}\frac{1}{\frac{1}{2\sqrt{x}}}=2\sqrt{8}=4\sqrt{2}$. If we assume the function is $\lim_{x\rightarrow 8}\frac{x - 8}{x - 8}=1$ (but this is a trivial case). Let's assume the correct one is $\lim_{x\rightarrow 8}\frac{x - 8}{x - 8}$ in a non - trivial sense of using L'Hopital's Rule. For $\lim_{x\rightarrow 1}\frac{x^{3}-5x^{2}+4}{x - 1}$, when $x = 1$, $x^{3}-5x^{2}+4=1 - 5+4=0$ and $x - 1=0$, so it is in the $\frac{0}{0}$ form.

Step3: Differentiate numerator and denominator for Question 2

Let $f(x)=x^{3}-5x^{2}+4$ and $g(x)=x - 1$. Then $f'(x)=3x^{2}-10x$ and $g'(x)=1$.

Step4: Evaluate the new limit

$\lim_{x\rightarrow 1}\frac{3x^{2}-10x}{1}=3\times1^{2}-10\times1=3 - 10=-7$.

Answer:

Question 1: No correct option shown with the above - assumed correct functions. Question 2: - 7