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  • Find limits of <img alt="\mathrm{\lim _{x \rightarrow 0} \frac{x \cos x-\sin x}{x^2 \sin x}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim%20_%7Bx%20%5Crightarrow%200%7D%

Find limits of \mathrm{\lim _{x \rightarrow 0} \frac{x \cos x-\sin x}{x^2 \sin x}}
 

Option: 1

-\frac{1}{3}


Option: 2

0


Option: 3

-1


Option: 4

3


Answers (1)

best_answer

Famously \mathrm{\lim _{x \rightarrow 0} \frac{\sin x}{x}=1} can be proved without such techniques, and implies \mathrm{\lim _{x \rightarrow 0} \frac{1-\cos x}{x^2}=\frac{1}{2}}. With a little more effort (e.g. by approximating a circular are as a parabola) you can also show \mathrm{\lim _{x \rightarrow 0} \frac{x-\sin x}{x^3}=\frac{1}{6}}. So \mathrm{\lim _{x \rightarrow 0} \frac{x-\sin x}{x^2 \sin x}=\frac{1}{6}} and
\mathrm{ \lim _{x \rightarrow 0}\left(\frac{\cos x-1}{x \sin x}+\frac{x-\sin x}{x^2 \sin x}\right)=-\frac{1}{2}+\frac{1}{6}=-\frac{1}{3} .}

Hence option 1 is correct.

Posted by

manish

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