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  • What is <img alt="\mathrm{\lim _{x \rightarrow \infty} x\left(\sqrt{x^2+2 x}-2 \sqrt{x^2+x}+x\right)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim%20_%7Bx%20%5Crigh

What is \mathrm{\lim _{x \rightarrow \infty} x\left(\sqrt{x^2+2 x}-2 \sqrt{x^2+x}+x\right)}

Option: 1

1


Option: 2

-\frac{1}{4}


Option: 3

2


Option: 4

3


Answers (1)

best_answer

Here is an elementary way by substitution \mathrm{x=\frac{1}{t}} and considering \mathrm{t \rightarrow 0^{+} } :

\begin{aligned} x\left(\sqrt{x^2+2 x}-2 \sqrt{x^2+x}+x\right) & \stackrel{x=\frac{1}{t}}{=} \frac{1}{t} \frac{\sqrt{1+2 t}-2 \sqrt{1+t}+1}{t} \\ \\& =\left(\frac{\sqrt{1+2 t}-1}{t^2}+2 \frac{1-\sqrt{1+t}}{t^2}\right) \\ \\& =\frac{2}{t(\sqrt{1+2 t}+1)}-\frac{2}{t(1+\sqrt{1+t})} \end{aligned}

 

                                                                  \begin{aligned} & =\frac{2}{(1+\sqrt{1+t})((\sqrt{1+2 t}+1))} \underbrace{\frac{\sqrt{1+t}-\sqrt{1+2 t}}{t}}_{=-\frac{1}{\sqrt{1+7+\sqrt{1+2}}}} \\ & \stackrel{t \rightarrow 0^{+}}{\longrightarrow} \frac{2}{4} \cdot \frac{-1}{2}=-\frac{1}{4} \end{aligned}

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