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  • Evaluate  Finding  <img alt="\mathrm{\lim _{x \rightarrow-\infty} \sqrt{x^2-5 x+1}-x}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim%20_%7Bx%20%5Crightarrow

Evaluate  Finding  \mathrm{\lim _{x \rightarrow-\infty} \sqrt{x^2-5 x+1}-x}

Option: 1

-5/2


Option: 2

1/2


Option: 3

3/5


Option: 4

1


Answers (1)

best_answer

Let   \mathrm{f(x)=\sqrt{x^2-5 x+1}-x}

Find   \mathrm{\lim _{x \rightarrow \infty} f(x)}

                                   \mathrm{\begin{aligned} & \lim _{x \rightarrow \infty} \sqrt{x^2-5 x+1}-x \\\\ & \lim _{x \rightarrow \infty} \frac{x^2-5 x+1-x^2}{\sqrt{x^2-5 x+1}+x} \\\\ & \lim _{x \rightarrow \infty} \frac{-5 x+1}{\sqrt{x^2-5 x+1}+x} \\\\ & \lim _{x \rightarrow \infty} \frac{\frac{-5 x+1}{x}}{\frac{\sqrt{x^2-5 x+1}+x}{x}} \\\\ & \lim _{x \rightarrow \infty} \frac{-5+\frac{1}{x}}{\sqrt{\frac{x^2}{x^2}-\frac{5}{x}+\frac{1}{x^2}}+1} \\ & \end{aligned}}

From here, I know that   \mathrm{\lim _{x \rightarrow \infty} \frac{1}{x}=0, \lim _{x \rightarrow \infty} \frac{5}{x}=0 \text { and } \lim _{x \rightarrow \infty} \frac{1}{x^2}=0} 

                                   \mathrm{\begin{gathered} \lim _{x \rightarrow \infty} \frac{-5}{\sqrt{1}+1} \\\\ \lim _{x \rightarrow \infty} \frac{-5}{2} \\\\ \lim _{x \rightarrow \infty} f(x)=\frac{-5}{2} \end{gathered}} 

Posted by

Shailly goel

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