evaluate the following limit: lim(x→+∞) (ln(2x + 4))/(4x + 4). enter -1 if your answer is -∞, enter 1 if…
evaluate the following limit: lim(x→+∞) (ln(2x + 4))/(4x + 4). enter -1 if your answer is -∞, enter 1 if your answer is ∞, and enter dne if the limit does not exist. limit =
Answer
Explanation:
Step1: Apply L'Hopital's Rule
As $x\to\infty$, we have the indeterminate form $\frac{\infty}{\infty}$. So, by L'Hopital's Rule, if $\lim_{x\to a}\frac{f(x)}{g(x)}$ is in the form $\frac{\infty}{\infty}$ or $\frac{0}{0}$, then $\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}$. Here, $f(x)=\ln(2x + 4)$ and $g(x)=4x + 4$. The derivative of $f(x)$ using the chain - rule is $f'(x)=\frac{2}{2x + 4}$, and the derivative of $g(x)$ is $g'(x)=4$.
Step2: Calculate the new limit
We now find $\lim_{x\to\infty}\frac{f'(x)}{g'(x)}=\lim_{x\to\infty}\frac{\frac{2}{2x + 4}}{4}=\lim_{x\to\infty}\frac{2}{4(2x + 4)}=\lim_{x\to\infty}\frac{2}{8x+16}$.
Step3: Evaluate the limit
As $x\to\infty$, the denominator $8x + 16\to\infty$. So, $\lim_{x\to\infty}\frac{2}{8x + 16}=0$.
Answer:
0