Question

use lhopitals rule to find the limit of $lim_{x \to infty} \frac{ln(x + 9)}{log_{2}x}$.
$lim_{x \to infty} \frac{ln(x + 9)}{log_{2}x} = square$

Explanation:

Step1: Apply L'Hôpital's Rule

L'Hôpital's Rule states that if \(\lim_{x
ightarrow a}\frac{f(x)}{g(x)}\) is in the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then \(\lim_{x
ightarrow a}\frac{f(x)}{g(x)}=\lim_{x
ightarrow a}\frac{f^{\prime}(x)}{g^{\prime}(x)}\).
Here, as \(x
ightarrow\infty\), \(\ln(x + 9)
ightarrow\infty\) and \(\log_{2}x=\frac{\ln x}{\ln 2}
ightarrow\infty\).
The derivative of \(y = \ln(x + 9)\) is \(y^{\prime}=\frac{1}{x+9}\) (using the chain - rule \((\ln u)^\prime=\frac{u^\prime}{u}\) with \(u=x + 9\)).
The derivative of \(y=\log_{2}x=\frac{\ln x}{\ln 2}\) is \(y^{\prime}=\frac{1}{x\ln 2}\) (using the formula \((\log_{a}x)^\prime=\frac{1}{x\ln a}\)).
So, \(\lim_{x
ightarrow\infty}\frac{\ln(x + 9)}{\log_{2}x}=\lim_{x
ightarrow\infty}\frac{\frac{1}{x + 9}}{\frac{1}{x\ln 2}}\).

Step2: Simplify the expression

\(\lim_{x
ightarrow\infty}\frac{\frac{1}{x + 9}}{\frac{1}{x\ln 2}}=\lim_{x
ightarrow\infty}\frac{x\ln 2}{x + 9}\).
Divide both the numerator and denominator by \(x\): \(\lim_{x
ightarrow\infty}\frac{\ln 2}{1+\frac{9}{x}}\).

Step3: Evaluate the limit

As \(x
ightarrow\infty\), \(\frac{9}{x}
ightarrow0\).
So, \(\lim_{x
ightarrow\infty}\frac{\ln 2}{1+\frac{9}{x}}=\ln 2\).

Answer:

\(\ln 2\)