Question

use lhôpitals rule to find the following limit.
lim (ln 3x - ln (x + 7))
x→∞

lim (ln 3x - ln (x + 7)) = (type an exact answer.)
x→∞

Explanation:

Step1: Use logarithmic property

Use $\ln a-\ln b=\ln\frac{a}{b}$. So, $\lim_{x
ightarrow\infty}(\ln3x - \ln(x + 7))=\lim_{x
ightarrow\infty}\ln\frac{3x}{x + 7}$.

Step2: Simplify the fraction inside the log

Divide numerator and denominator by $x$: $\frac{3x}{x + 7}=\frac{3}{1+\frac{7}{x}}$.

Step3: Find the limit of the fraction

As $x
ightarrow\infty$, $\lim_{x
ightarrow\infty}\frac{7}{x}=0$. Then $\lim_{x
ightarrow\infty}\frac{3}{1+\frac{7}{x}}=\frac{3}{1 + 0}=3$.

Step4: Find the limit of the log - function

Since $y = \ln u$ is continuous at $u = 3$, $\lim_{x
ightarrow\infty}\ln\frac{3x}{x + 7}=\ln(\lim_{x
ightarrow\infty}\frac{3x}{x + 7})$.

Answer:

$\ln3$