Use l'H^opital's rule to find the limit.
-$\lim _ { x \rightarrow \infty } \left( \sqrt { x ^ { 2 } + 6 x } - x \right)$
A) - 3
B) 0
C) 3
D) 6

Question

Quizplus · Accepted Answer

First, we can simplify the expression by rationalizing the numerator:
\begin{align*}
\sqrt{x^2+6x}-x &= \frac{(\sqrt{x^2+6x}-x)(\sqrt{x^2+6x}+x)}{\sqrt{x^2+6x}+x} \
&= \frac{(x^2+6x-x^2)}{\sqrt{x^2+6x}+x} \
&= \frac{6x}{\sqrt{x^2+6x}+x}
\end{align*}
Then, we can apply l'H^opital's rule by taking the derivative of the numerator and denominator with respect to x:
\begin{align*}
\lim_{xightarrow\infty} \frac{6x}{\sqrt{x^2+6x}+x} &= \lim_{xightarrow\infty} \frac{6}{\frac{1}{2\sqrt{x^2+6x}}+1} \
&= \frac{6}{\frac{1}{2\cdot\infty}+1} \
&= 3
\end{align*}
Therefore, the answer is (C) 3.

Use L'H^opital's Rule to Find the Limit lim⁡x→∞(x2+6x−x)\lim _ { x \rightarrow \infty } \left( \sqrt { x ^ { 2 } + 6 x } - x \right)limx→∞​(x2+6x​−x)

Use L'H^opital's Rule to Find the Limit $\lim _ { x \rightarrow \infty } \left( \sqrt { x ^ { 2 } + 6 x } - x \right)$