Questions: For the function (f(x)) given below, where (x>-4), evaluate (lim x rightarrow infty f(x)). [f(x)=ln (2 x+8)-ln (6 x^2+7)] If the function increases without bound, you should enter (infty). If the function decreases without bound, you should enter (-infty). If the function does not approach a finite limit nor (pm infty) as (x rightarrow pm infty), you should enter (varnothing). Provide your answer below: [lim x rightarrow infty f(x)=]

For the function (f(x)) given below, where (x>-4), evaluate (lim x rightarrow infty f(x)).
[f(x)=ln (2 x+8)-ln (6 x^2+7)]

If the function increases without bound, you should enter (infty). If the function decreases without bound, you should enter (-infty). If the function does not approach a finite limit nor (pm infty) as (x rightarrow pm infty), you should enter (varnothing).

Provide your answer below:
[lim x rightarrow infty f(x)=]

Transcript text: For the function $f(x)$ given below, where $x>-4$, evaluate $\lim _{x \rightarrow \infty} f(x)$. \[ f(x)=\ln (2 x+8)-\ln \left(6 x^{2}+7\right) \] If the function increases without bound, you should enter $\infty$. If the function decreases without bound, you should enter $-\infty$. If the function does not approach a finite limit nor $\pm \infty$ as $x \rightarrow \pm \infty$, you should enter $\varnothing$. Provide your answer below: \[ \lim _{x \rightarrow \infty} f(x)= \]

Solution

Solution Steps

To evaluate the limit of the given function as $ x $ approaches infinity, we can simplify the expression using properties of logarithms. Specifically, we can use the property $\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right)$. Then, we analyze the behavior of the resulting expression as $ x $ approaches infinity.

Step 1: Simplifying the Function

We start with the function given by

\[ f(x) = \ln(2x + 8) - \ln(6x^2 + 7). \]

Using the property of logarithms, we can combine the two logarithmic terms:

\[ f(x) = \ln\left(\frac{2x + 8}{6x^2 + 7}\right). \]

Step 2: Analyzing the Limit

Next, we need to evaluate the limit as $ x $ approaches infinity:

\[ \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \ln\left(\frac{2x + 8}{6x^2 + 7}\right). \]

As $ x $ becomes very large, the dominant terms in the numerator and denominator are $ 2x $ and $ 6x^2 $, respectively. Thus, we can simplify the fraction:

\[ \frac{2x + 8}{6x^2 + 7} \approx \frac{2x}{6x^2} = \frac{2}{6x} = \frac{1}{3x}. \]

Step 3: Evaluating the Limit of the Logarithm

Now we evaluate the limit of the logarithm:

\[ \lim_{x \to \infty} \ln\left(\frac{1}{3x}\right) = \lim_{x \to \infty} \left(\ln(1) - \ln(3x)\right) = 0 - \lim_{x \to \infty} \ln(3x). \]

Since $ \ln(3x) $ approaches infinity as $ x $ approaches infinity, we have:

\[ \lim_{x \to \infty} f(x) = -\infty. \]