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  • Evaluating: <img alt="\mathrm{ \lim _{x \rightarrow 0}\left(\frac{(\sin x)^2-x^2}{x^4}\right) }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%20%5Clim%20_%7Bx%20%5Cri

Evaluating:

\mathrm{ \lim _{x \rightarrow 0}\left(\frac{(\sin x)^2-x^2}{x^4}\right) }

Option: 1

\frac{1}{2}


Option: 2

1


Option: 3

-\frac{1}{3}


Option: 4

0


Answers (1)

best_answer

                                     \mathrm{\lim _{x \rightarrow 0} \frac{(\sin x)^2-x^2}{x^4}}

                                   \mathrm{=\lim _{x \rightarrow 0} \frac{\frac{1-\cos 2 x}{2}-x^2}{x^4}}

                                   \mathrm{=\frac{1}{2} \lim _{x \rightarrow 0} \frac{1-\cos 2 x-2 x^2}{x^4}}

Using L'Hospital's rule for \frac{0}{0} form four times, one should get

                                   \mathrm{=\frac{1}{2} \lim _{x \rightarrow 0} \frac{-2^4 \cos 2 x}{4 \cdot 3 \cdot 2 \cdot 1}}

                                  \mathrm{=-\frac{16 \cos 0}{48}=-\frac{1}{3}}

Posted by

HARSH KANKARIA

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