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  • Evaluating: <img alt="\mathrm{\quad \lim _{x \rightarrow 0}\left(\frac{1}{\sin x}-\frac{1}{\tan x}\right)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cquad%20%5Cl

Evaluating:

\mathrm{\quad \lim _{x \rightarrow 0}\left(\frac{1}{\sin x}-\frac{1}{\tan x}\right)}

Option: 1

0


Option: 2

1


Option: 3

10


Option: 4

5


Answers (1)

best_answer

\mathrm{\lim _{x \rightarrow 0}\left(\frac{1}{\sin x}-\frac{1}{\tan x}\right)=\lim _{x \rightarrow 0}\left(\frac{1}{\sin x}-\frac{\cos x}{\sin x}\right)}

                                            \mathrm{=\lim _{x \rightarrow 0}\left(\frac{1-\cos x}{\sin x}\right)}

                                           \mathrm{=\lim _{x \rightarrow 0}\left(\frac{1-\cos x}{\sin x} \cdot \frac{1+\cos x}{1+\cos x}\right)}

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

                                         \mathrm{=\lim _{x \rightarrow 0}\left(\frac{\sin ^2 x}{\sin x(1+\cos x)}\right)}

                                         \mathrm{=\lim _{x \rightarrow 0}\left(\frac{\sin x}{1+\cos x}\right)}

                                        \mathrm{=\frac{\sin 0}{1+\cos 0}}

                                        \mathrm{=0}

Posted by

SANGALDEEP SINGH

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