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  • Evaluate the limit:<img alt="\lim_{x\rightarrow \infty }\left ( \sqrt{2x^{2}+5x} -x\right )" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Clim_%7Bx%5Crightarrow%20%5Cinfty%20%7D%5Cleft%20

Evaluate the limit:\lim_{x\rightarrow \infty }\left ( \sqrt{2x^{2}+5x} -x\right )

Option: 1

0


Option: 2

1


Option: 3

5


Option: 4

\infty


Answers (1)

best_answer

For solving the limits of the type ∞ - ∞, we simply rationalize the given limit, that is multiplying and dividing the limit by its additive inverse.

Additive inverse of \lim_{x\rightarrow \infty }\left ( \sqrt{2x^{2}+5x} -x\right ) is \lim_{x\rightarrow \infty }\left ( \sqrt{2x^{2}+5x} +x\right )

Hence,

\begin{aligned} & \lim _{x \rightarrow \infty}\left(\sqrt{2 x^2+5 x}+x\right) \\ = & \lim _{x \rightarrow \infty} \frac{\left(\sqrt{2 x^2+5 x}-x\right)\left(\sqrt{2 x^2+5 x}+x\right)}{\left(\sqrt{2 x^2+5 x}+x\right)} \\ = & \lim _{x \rightarrow \infty} \frac{2 x^2+5 x-x^2}{\left(\sqrt{2 x^2+5 x}+x\right)} \\ = & \lim _{x \rightarrow \infty} \frac{x^2+5 x}{\left(\sqrt{2 x^2+5 x}+x\right)} \\ = & \lim _{x \rightarrow \infty} \frac{x^2\left(1+\frac{5}{x}\right)}{\left(x^2\left(\sqrt{\frac{2}{x^2}+\frac{5}{x^3}}+\frac{1}{x}\right)\right)} \\ = & \frac{1}{0} \\ = & \infty \end{aligned}

Posted by

Ritika Kankaria

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