Question

use lhôpitals rule to find the limit of $limlimits_{x\toinfty}\frac{ln(x + 9)}{log_3x}$
$limlimits_{x\toinfty}\frac{ln(x + 9)}{log_3x}=square$

Explanation:

Step1: Recall L'Hôpital's Rule and Logarithm Properties

L'Hôpital's Rule: If \(\lim_{x
ightarrow a}\frac{f(x)}{g(x)}\) is in the \(\frac{\infty}{\infty}\) or \(\frac{0}{0}\) form, then \(\lim_{x
ightarrow a}\frac{f(x)}{g(x)}=\lim_{x
ightarrow a}\frac{f^{\prime}(x)}{g^{\prime}(x)}\). Also, \(\log_{3}x=\frac{\ln x}{\ln 3}\). So, \(\lim_{x
ightarrow\infty}\frac{\ln(x + 9)}{\log_{3}x}=\lim_{x
ightarrow\infty}\frac{\ln(x + 9)}{\frac{\ln x}{\ln 3}}=\ln 3\lim_{x
ightarrow\infty}\frac{\ln(x + 9)}{\ln x}\). As \(x
ightarrow\infty\), \(\frac{\ln(x + 9)}{\ln x}\) is in the \(\frac{\infty}{\infty}\) form.

Step2: Differentiate Numerator and Denominator

Differentiate \(y = \ln(x + 9)\) using the chain - rule: \(y^{\prime}=\frac{1}{x + 9}\). Differentiate \(y=\ln x\): \(y^{\prime}=\frac{1}{x}\).

Step3: Apply L'Hôpital's Rule

By L'Hôpital's Rule, \(\ln 3\lim_{x
ightarrow\infty}\frac{\ln(x + 9)}{\ln x}=\ln 3\lim_{x
ightarrow\infty}\frac{\frac{1}{x + 9}}{\frac{1}{x}}\). Simplify \(\frac{\frac{1}{x + 9}}{\frac{1}{x}}=\frac{x}{x + 9}\).

Step4: Simplify the Limit

\(\lim_{x
ightarrow\infty}\frac{x}{x + 9}=\lim_{x
ightarrow\infty}\frac{x}{x(1+\frac{9}{x})}\). Cancel out \(x\) (since \(x
eq0\) as \(x
ightarrow\infty\)), we get \(\lim_{x
ightarrow\infty}\frac{1}{1+\frac{9}{x}}\). As \(x
ightarrow\infty\), \(\frac{9}{x}
ightarrow0\).

Answer:

\(\ln 3\)