Evaluate the following limits: _ x [x(x+1-x)] - Vidyadip

Question

Evaluate the following limits: $\lim _{x \rightarrow \infty}[\sqrt{x}(\sqrt{x+1}-\sqrt{x})]$

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Answer

$\lim _{x \rightarrow 0}[\sqrt{x}(\sqrt{x+1}-\sqrt{x})]$
$=\lim _{x \rightarrow 0} \sqrt{x}(\sqrt{x+1}-\sqrt{x}) \times \frac{(\sqrt{x+1}+\sqrt{x})}{\sqrt{x+1}+\sqrt{x}}
...[$By rationalization$]$
$=\lim _{x \rightarrow 0} \frac{\sqrt{x}(x+1-x)}{\sqrt{x+1}+\sqrt{x}}$
$=\lim _{x \rightarrow 0} \frac{\sqrt{x}}{\sqrt{x+1}+\sqrt{x}}$
$=\lim _{x \rightarrow \infty} \frac{1}{\frac{\sqrt{x+1}}{\sqrt{x}}+1} \quad \cdots\left[\begin{array}{l}
\text { Divide numerator and } \\\text { denominator by } \sqrt{x}\end{array}\right]$
$=\lim _{x \rightarrow \infty} \frac{1}{\sqrt{\frac{x+1}{x}+1}}$
$=\frac{1}{\lim _{x \rightarrow \infty}\left(\sqrt{1+\frac{1}{x}}+1\right)}$
$=\frac{1}{\sqrt{1+0}+1} \ldots\left[\lim _{x \rightarrow \infty} \frac{1}{x^k}=0, k>0\right]$
$=\frac{1}{1+1}$
$=\frac{1}{2}$

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