Certainly! Let's solve the given limit step-by-step.
We start with the expression for the limit:
[tex]\[
\lim_{x \to \infty} \frac{16x - 8x}{4x^2 + 2x}
\][/tex]
### Step 1: Simplify the Expression
First, we simplify the numerator [tex]\(16x - 8x\)[/tex]:
[tex]\[
16x - 8x = 8x
\][/tex]
So, the expression becomes:
[tex]\[
\frac{8x}{4x^2 + 2x}
\][/tex]
### Step 2: Factor the Denominator
Next, let's factor the denominator [tex]\(4x^2 + 2x\)[/tex]:
[tex]\[
4x^2 + 2x = 2x(2x + 1)
\][/tex]
Thus, the expression now looks like this:
[tex]\[
\frac{8x}{2x(2x + 1)}
\][/tex]
### Step 3: Simplify the Fraction
We can now simplify the fraction by canceling out [tex]\(2x\)[/tex] in the numerator and the denominator:
[tex]\[
\frac{8x}{2x(2x + 1)} = \frac{8}{2(2x + 1)}
\][/tex]
Simplifying further, we get:
[tex]\[
\frac{8}{2(2x + 1)} = \frac{4}{2x + 1}
\][/tex]
### Step 4: Take the Limit as [tex]\(x\)[/tex] Approaches Infinity
Now, we need to compute the limit of [tex]\(\frac{4}{2x + 1}\)[/tex] as [tex]\(x\)[/tex] approaches infinity:
[tex]\[
\lim_{x \to \infty} \frac{4}{2x + 1}
\][/tex]
As [tex]\(x\)[/tex] grows larger and larger, [tex]\(2x + 1\)[/tex] also becomes extremely large. Therefore, the fraction [tex]\(\frac{4}{2x + 1}\)[/tex] becomes very small. More formally:
[tex]\[
\lim_{x \to \infty} \frac{4}{2x + 1} = 0
\][/tex]
### Final Answer
So, the limit is:
[tex]\[
\boxed{0}
\][/tex]
