#### explain solution RD Sharma class 12 Chapter 8 Continuity exercise 8.1 question 13

$a=\frac{1}{2}$

Hint:

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 to the value of the right hand or left hand limit at that point.

Solution:

Given,

$f(x)=\left(\begin{array}{l} a \sin \frac{\pi}{2}(x+1), \text { if } x \leq 0 \\\\ \frac{\tan x-\sin x}{x^{3}}, \text { if } x>0 \end{array}\right)$

We observe

[LHL at $x=0$ ]

$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(-h)=\lim _{h \rightarrow 0} a \sin \frac{\pi}{2}(-h+1)=a \sin \frac{\pi}{2}=a$               $\left[\because \sin \frac{\pi}{2}=1\right]$

[RHL 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} \frac{\tanh -\sinh }{h^{3}} \end{aligned}                                                                                                 $\left[\begin{array}{l} \because \frac{\sin x}{\cos x}=\tan x \\\\ \because 1-\cos x=2 \sin ^{2} \frac{x}{2} \end{array}\right]$

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{\frac{\sinh }{\cosh }-\sinh }{h^{3}} \\\\ &=\lim _{h \rightarrow 0} \frac{(1-\cosh ) \tanh }{h^{3}} \end{aligned}

$=\lim _{h \rightarrow 0} \frac{2\left(\sin ^{2} \frac{h}{2}\right) \tanh }{\frac{4 h^{2}}{4} \times h}$                   [Multiplying and dividing the denominator by $4$ ]

$=\frac{2}{4} \lim _{h \rightarrow 0} \frac{\left(\sin ^{2} \frac{h}{2}\right) \tanh }{\frac{h^{2}}{4} \times h}$                                                                          $\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$

$=\frac{1}{2} \lim _{h \rightarrow 0} \frac{\left(\sin \frac{h}{2}\right)}{\frac{h}{2}} \lim _{h \rightarrow 0} \frac{\tanh }{h}$                                                                    $\left[\because \lim _{x \rightarrow 0} \frac{\tan x}{x}=1\right]$

\begin{aligned} &\lim _{x \rightarrow 0^{+}} f(x)=\frac{1}{2} \times 1 \times 1 \\\\ &\lim _{x \rightarrow 0^{+}} f(x)=\frac{1}{2} \end{aligned}\begin{aligned} &\lim _{x \rightarrow 0^{+}} f(x)=\frac{1}{2} \times 1 \times 1 \\ &\lim _{x \rightarrow 0^{+}} f(x)=\frac{1}{2} \end{aligned}

If $f\left ( x \right )$ is continuous at $x=0$ , then

\begin{aligned} &\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{-}} f(x) \\\\ &a=\frac{1}{2} \end{aligned}