Evaluate the following limits: _ x [x^4+4 x^2-x^2 ] - Vidyadip

Question

Evaluate the following limits: $\lim _{x \rightarrow \infty}\left[\sqrt{x^4+4 x^2}-x^2\right]$

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Answer

$\lim _{x \rightarrow \infty}\left[\sqrt{x^4+4 x^2}-x^2\right]$
$=\lim _{x \rightarrow \infty} \frac{\left(\sqrt{x^4+4 x^2}-x^2\right)\left(\sqrt{x^4+4 x^2}+x^2\right)}{\sqrt{x^4+4 x^2}+x^2}$
$[$By rationalizatio$]$
$=\lim _{x \rightarrow \infty} \frac{x^4+4 x^2-x^4}{\sqrt{x^4+4 x^2}+x^2}$
$=\lim _{x \rightarrow \infty} \frac{4 x^2}{\sqrt{x^4+4 x^2}+x^2}$
$=\lim _{x \rightarrow \infty} \frac{\frac{4 x^2}{x^2}}{\frac{\sqrt{x^4+4 x^2}+x^2}{x^2}} \cdots\left[\begin{array}{l}
\text { Divide numerator and } \\ \text { denominator by } x^2 \end{array}\right]$
$=\lim _{x \rightarrow \infty} \frac{\frac{4 x^2}{x^2}}{\sqrt{\frac{x^4+4 x^2}{x^4}+1}}$
$=\frac{\lim _{x \rightarrow \infty} 4}{\lim _{x \rightarrow \infty}\left(\sqrt{1+\frac{4}{x^2}}+1\right)}$
$=\frac{4}{\sqrt{1+0}+1} \ldots\left[\lim _{x \rightarrow \infty} \frac{1}{x^{ k }}=0, k >0\right]$
$=\frac{4}{2}$
$=2$

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