Question

4.6 limits at infinity and horizontal asymp - (1 point) evaluate $lim_{x \to +infty}(sqrt{16x^{2}+x}-4x)$. enter inf for $infty$, -inf for $-infty$, and dne if the limit does not exist. limit = $infty$ note: you are in the reduced scoring period. all work counts for 85% of the original. preview my answers submit answers you have attempted this problem 1 time. your overall recorded score is 0%. you have unlimited attempts remaining.

Explanation:

Step1: Rationalize the expression

Multiply and divide by $\sqrt{16x^{2}+x}+4x$.
\[

$$\begin{align*} &\lim_{x ightarrow+\infty}(\sqrt{16x^{2}+x}-4x)\times\frac{\sqrt{16x^{2}+x}+4x}{\sqrt{16x^{2}+x}+4x}\\ =&\lim_{x ightarrow+\infty}\frac{(\sqrt{16x^{2}+x})^{2}-(4x)^{2}}{\sqrt{16x^{2}+x}+4x}\\ =&\lim_{x ightarrow+\infty}\frac{16x^{2}+x - 16x^{2}}{\sqrt{16x^{2}+x}+4x}\\ =&\lim_{x ightarrow+\infty}\frac{x}{\sqrt{16x^{2}+x}+4x} \end{align*}$$

\]

Step2: Divide numerator and denominator by $x$

Since $x>0$ as $x
ightarrow+\infty$, we have:
\[

$$\begin{align*} &\lim_{x ightarrow+\infty}\frac{x}{\sqrt{16x^{2}+x}+4x}\\ =&\lim_{x ightarrow+\infty}\frac{x/x}{\sqrt{16x^{2}+x}/x + 4x/x}\\ =&\lim_{x ightarrow+\infty}\frac{1}{\sqrt{16+\frac{1}{x}}+4} \end{align*}$$

\]

Step3: Evaluate the limit

As $x
ightarrow+\infty$, $\frac{1}{x}
ightarrow0$.
\[

$$\begin{align*} &\lim_{x ightarrow+\infty}\frac{1}{\sqrt{16+\frac{1}{x}}+4}\\ =&\frac{1}{\sqrt{16 + 0}+4}\\ =&\frac{1}{4 + 4}\\ =&\frac{1}{8} \end{align*}$$

\]

Answer:

$\frac{1}{8}$