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  • Evaluate <img alt="\mathrm{\lim _{x \rightarrow \pi / 2}(\frac{x}{tan(x)}-x)/(x-\frac{\pi }{2})}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim%20_%7Bx%20%5Crightarro

Evaluate \mathrm{\lim _{x \rightarrow \pi / 2}(\frac{x}{tan(x)}-x)/(x-\frac{\pi }{2})}

Option: 1

\mathrm{\frac{5\pi }{2}}


Option: 2

\mathrm{\frac{3\pi}{2}}


Option: 3

\mathrm{-\frac{\pi}{2}}


Option: 4

\mathrm{\frac{\pi}{2}}


Answers (1)

best_answer

Note that \mathrm{\lim _{x \rightarrow \pi / 2} \arctan \left(\frac{x}{\tan (x)-x}\right)=0}. Then thios just the limit definition of the derivative :

\mathrm{\begin{gathered} \lim _{x \rightarrow \pi / 2} \frac{\arctan \left(\frac{x}{\tan (x)-x}\right)-0}{x-\pi / 2} \\ =\lim _{x \rightarrow \pi / 2} \frac{d}{d x} \arctan \left(\frac{x}{\tan (x)-x}\right) \end{gathered}}

A tendious chain-rule calculation produces :

\mathrm{=\lim _{x \rightarrow \pi / 2} \frac{-x \sec ^2(x)+\tan (x)}{\tan ^2(x)+2 x^2-2 x \tan (x)}}

Multiply through by \mathrm{cos ^{2}(x)} to evaluate the limit .

\mathrm{=\lim _{x \rightarrow \pi / 2} \frac{-x+\sin (x) \cos (x)}{\sin ^2(x)+2 x^2 \cos ^2(x)-2 x \sin (x) \cos (x)}=\frac{-\pi / 2+0}{1+0+0}=\frac{-\pi}{2}}

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chirag

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