Question

use lhopitals rule to find the limit. note that in this problem, neither algebraic simplification nor the theorem for limits of rational functions at infinity provides an alternative to lhopitals rule.
lim_{x \to 1} \frac{8x - 8}{ln x^4}
select the correct choice below and, if necessary, fill in the answer box to complete your choice.
oa. ( lim_{x \to 1} \frac{8x - 8}{ln x^4} = ) (simplify your answer.)
ob. the limit does not exist.

Explanation:

Step1: Check the form of the limit

When \(x = 1\), the numerator \(8x-8=8\times1 - 8=0\), and the denominator \(\ln x=\ln1 = 0\). So, we have the \(\frac{0}{0}\) form, and we can apply L'Hopital's rule.
L'Hopital's rule states that if \(\lim_{x
ightarrow a}\frac{f(x)}{g(x)}\) is in the \(\frac{0}{0}\) or \(\frac{\pm\infty}{\pm\infty}\) form, then \(\lim_{x
ightarrow a}\frac{f(x)}{g(x)}=\lim_{x
ightarrow a}\frac{f^{\prime}(x)}{g^{\prime}(x)}\)

Differentiate the numerator \(f(x)=8x - 8\), using the power rule \((x^n)^\prime=nx^{n - 1}\), we get \(f^{\prime}(x)=8\)

Differentiate the denominator \(g(x)=\ln x\), using the formula \((\ln x)^\prime=\frac{1}{x}\), we get \(g^{\prime}(x)=\frac{1}{x}\)

So, \(\lim_{x
ightarrow1}\frac{8x - 8}{\ln x}=\lim_{x
ightarrow1}\frac{8}{\frac{1}{x}}\)

Step2: Evaluate the new limit

\(\lim_{x
ightarrow1}\frac{8}{\frac{1}{x}}=\lim_{x
ightarrow1}(8x)\)

Substitute \(x = 1\) into \(8x\), we get \(8\times1=8\)

Answer:

A. \(\lim_{x
ightarrow1}\frac{8x - 8}{\ln x}=8\)