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  • Evaluate   <img alt="\mathrm{\lim _{x \rightarrow 0} \frac{\sin ^{-1}(x)}{\sin (3 x)}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim%20_%7Bx%20%5Crightarrow%200

Evaluate   \mathrm{\lim _{x \rightarrow 0} \frac{\sin ^{-1}(x)}{\sin (3 x)}}

Option: 1

0


Option: 2

1/3


Option: 3

2


Option: 4

5


Answers (1)

best_answer

                                            \mathrm{\lim _{x \rightarrow 0} \frac{\sin ^{-1}(x)}{\sin (3 x)}=}

We substitute X by Sin(t). This is possible because sin is a strict increasing continuous function 

near 0 and sin(0)=0 and get 

                                \mathrm{\begin{gathered} =\lim _{t \rightarrow 0} \frac{\sin ^{-1}(\sin (t))}{\sin (3 \sin (t))}= \\\\ =\lim _{t \rightarrow 0} \frac{t}{\sin (3 \sin (t))}= \\\\ =\lim _{t \rightarrow 0} \frac{t}{\sin (t)} \frac{\sin (t)}{\sin (3 \sin (t))}= \\\\ =\lim _{t \rightarrow 0} \frac{t}{\sin (t)} \lim _{t \rightarrow 0} \frac{\sin (t)}{\sin (3 \sin (t))}= \\\\ =\lim _{t \rightarrow 0} \frac{t}{\sin (t)} \frac{1}{3} \lim _{t \rightarrow 0} \frac{3 \sin (t)}{\sin (3 \sin (t))}= \end{gathered}}

and if we substitute 3 \mathrm{\sin t } by \mathrm{ u} then we get

                                 \mathrm{\begin{gathered} =\lim _{t \rightarrow 0} \frac{t}{\sin (t)} \frac{1}{3} \lim _{u \rightarrow 0} \frac{u}{\sin (u)}= \\\\ =\frac{1}{3} \end{gathered}} 

 

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