Question

use lhôpitals rule to find the limit of $lim_{x
ightarrowinfty}\frac{ln(x + 9)}{log_4 x}$. $lim_{x
ightarrowinfty}\frac{ln(x + 9)}{log_4 x}=square$

Explanation:

Step1: Check form of limit

As $x
ightarrow\infty$, $\ln(x + 9)
ightarrow\infty$ and $\log_4x
ightarrow\infty$, so it's in $\frac{\infty}{\infty}$ form.

Step2: Differentiate numerator and denominator

The derivative of $y=\ln(x + 9)$ using the chain - rule is $y^\prime=\frac{1}{x + 9}$. The derivative of $y = \log_4x=\frac{\ln x}{\ln 4}$, and its derivative is $y^\prime=\frac{1}{x\ln 4}$ by the formula $\frac{d}{dx}(\ln x)=\frac{1}{x}$.

Step3: Find new limit

We now find $\lim_{x
ightarrow\infty}\frac{\frac{1}{x + 9}}{\frac{1}{x\ln 4}}=\lim_{x
ightarrow\infty}\frac{x\ln 4}{x + 9}$.

Step4: Simplify new limit

Divide both numerator and denominator by $x$: $\lim_{x
ightarrow\infty}\frac{\ln 4}{1+\frac{9}{x}}$.

Step5: Evaluate limit

As $x
ightarrow\infty$, $\frac{9}{x}
ightarrow0$. So $\lim_{x
ightarrow\infty}\frac{\ln 4}{1+\frac{9}{x}}=\ln 4$.

Answer:

$\ln 4$