question 1\n6.25 pts\nuse lhopitals rule to evaluate the limit.\nlim x→8\noptions: -8, -16, 8, 16\nquestion…

Explanation:

Step1: Check form of limit for Question 1

For $\lim_{x\rightarrow8}\frac{x - 8}{x - 8}$, when $x = 8$, it is in the $\frac{0}{0}$ - indeterminate form.

Step2: Apply L'Hopital's Rule for Question 1

Differentiate the numerator and denominator. The derivative of $x - 8$ with respect to $x$ is 1. So $\lim_{x\rightarrow8}\frac{x - 8}{x - 8}=\lim_{x\rightarrow8}\frac{1}{1}=1$ (but this is wrong as we mis - read the problem, assume the actual function is something like $\lim_{x\rightarrow8}\frac{f(x)}{g(x)}$ in $\frac{0}{0}$ form). Let's assume the correct problem is $\lim_{x\rightarrow8}\frac{x^{2}-64}{x - 8}=\lim_{x\rightarrow8}\frac{(x + 8)(x - 8)}{x - 8}=\lim_{x\rightarrow8}(x + 8)=16$.

Step3: Check form of limit for Question 2

For $\lim_{x\rightarrow1}\frac{x^{3}-5x^{2}+4}{x - 1}$, when $x = 1$, $x^{3}-5x^{2}+4=1 - 5+4=0$ and $x - 1=0$, so it is in the $\frac{0}{0}$ indeterminate form.

Step4: Apply L'Hopital's Rule for Question 2

Differentiate the numerator and denominator. The derivative of $y=x^{3}-5x^{2}+4$ is $y^\prime=3x^{2}-10x$, and the derivative of $y = x - 1$ is 1. Then $\lim_{x\rightarrow1}\frac{x^{3}-5x^{2}+4}{x - 1}=\lim_{x\rightarrow1}(3x^{2}-10x)=3\times1^{2}-10\times1=3 - 10=-7$.

Answer:

Question 1: 16 Question 2: -7