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Please Solve RD Sharma Class 12 Chapter 8 Continuity Exercise 8.1 Question 12 Maths Textbook Solution.

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Continuous function must be defined at a point, limit must exist at the point and the value of the function at that point must equal the value of the right hand and left hand limit at that point.



f(x)=\left(\begin{array}{cl} \frac{\sin 3 x}{\tan 2 x} & , \text { if } x<0 \\\\ \frac{3}{2} & ,\text { if } x=0 \\\\ \frac{\log (1+3 x)}{e^{2 x}-1}&, \text { if } x> 0\\\\ \end{array}\right)

We observe

[LHL at  x=0 ]

\begin{aligned} &\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h) \\\\ &=\lim _{h \rightarrow 0} f(-h) \\\\ &=\lim _{h \rightarrow 0}\left(\frac{\sin 3(-h)}{\tan 2(-h)}\right) \end{aligned}

=\lim _{h \rightarrow 0}\left(\frac{\sin 3 h}{\tan 2 h}\right) \Rightarrow \lim _{h \rightarrow 0}\left(\frac{\frac{3 \sin 3 h}{3h}}{\frac{2 \tan 2 h}{2 h}}\right)

\begin{aligned} &=\frac{\lim _{h \rightarrow 0}\left(\frac{3 \sin 3 h}{3 h}\right)}{\lim _{h \rightarrow 0}\left(\frac{2 \tan 2 h}{2 h}\right)} \Rightarrow \frac{3 \lim _{h \rightarrow 0}\left(\frac{\sin 3 h}{3 h}\right)}{2 \lim _{h \rightarrow 0}\left(\frac{\tan 2 h}{2 h}\right)} \\\\ \end{aligned}                                                                \left[\begin{array}{l} \because \sin (-x)=-\sin x \\ \because \tan (-x)=-\tan x \end{array}\right]

=\frac{3 \times 1}{2 \times 1}=\frac{3}{2}   

                                                                                                                                                         \left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]

And    f\left ( 0 \right )=\frac{3}{2}

\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{-}} f(x)=f(0)

Thus, f\left ( x \right )  is continuous at  x=0.



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