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  • Using the power law for limits evaluate <img alt="\lim_{x\rightarrow -2}\left ( \frac{x^{3}-2}{x^{3}+14x^{2}+2} \right )^{2}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Clim_%7Bx%5

Using the power law for limits evaluate \lim_{x\rightarrow -2}\left ( \frac{x^{3}-2}{x^{3}+14x^{2}+2} \right )^{2}

Option: 1

\frac{1}{5}


Option: 2

\pm \frac{1}{5}


Option: 3

\frac{1}{5}


Option: 4

\frac{1}{25}


Answers (1)

best_answer

Note that the “Power law for limits” states that \lim_{x\rightarrow a}\left ( f\left ( x \right ) \right )^{n}= \left ( \lim_{x\rightarrow a}f\left ( x \right ) \right )^{n} for n= +ve integer.
 

Now, apply the “Power law for limits” to evaluate the provided limit in the following way.

\lim_{x\rightarrow -2}\left ( \frac{x^{3}-2}{x^{3}+14x^{2}+2} \right )^{2}

= \left ( \lim_{x\rightarrow -2} \frac{x^{3}-2}{x^{3}+14x^{2}+2} \right )^{2}
= \left [ \frac{\left ( -2 \right )^{3}-2}{\left ( -2 \right )^{3}+14\times\left ( -2 \right )^{2}+2} \right ]^{2}
= \left [\frac{ -8-2}{-8+14\times4+2}\right ]^{2}
=\left ( \frac{-10}{50} \right )^{2}
=\left ( -\frac{1}{5} \right )^{2}
=\frac{1}{25}

 

Posted by

Ritika Harsh

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