Find the limit.
-$$\lim _ { x \rightarrow } \sqrt { x ^ { 2 } + 18 x } - x$$
A) 9
B) 0
C) 18
D) $\infty$

Question

Quizplus · Accepted Answer

To find the limit, we can multiply and divide by the conjugate to simplify the expression: $$\lim _ { x \rightarrow \infty } \sqrt { x ^ { 2 } + 18 x } - x = \lim _ { x \rightarrow \infty } \frac{(\sqrt { x ^ { 2 } + 18 x } - x)(\sqrt { x ^ { 2 } + 18 x } + x)}{\sqrt { x ^ { 2 } + 18 x } + x}$$ This simplifies to $$\lim _ { x \rightarrow \infty } \frac{x^2 + 18x - x^2}{\sqrt { x ^ { 2 } + 18 x } + x} = \lim _ { x \rightarrow \infty } \frac{18x}{x(\sqrt { 1 + \frac{18}{x} } + 1)}$$ Simplifying further, we get $$\lim _ { x \rightarrow \infty } \frac{18}{\sqrt { 1 + \frac{18}{x} } + 1} = \frac{18}{\sqrt{1 + 0} + 1} = \frac{18}{2} = 9.$$

Find the Limit lim⁡x→x2+18x−x\lim _ { x \rightarrow } \sqrt { x ^ { 2 } + 18 x } - xx→lim​x2+18x​−x

Find the Limit $\lim _ { x \rightarrow } \sqrt { x ^ { 2 } + 18 x } - x$