#### Provide Solutio for RD Sharma Class 12 Chapter Continuity Exercise 8.1 Question 26

$a=\frac{-3}{2}, b \neq 0, c=\frac{1}{2} ; b \in R-\{0\}$

Hint:

$f\left ( x \right )$  must be defined. The limit of the $f\left ( x \right )$ approaches the value $x$ must exist.

Given:

$f(x)=\left\{\begin{array}{c} \frac{\sin (a+1) x+\sin x}{x}, \text { for } x<0 \\\\ c \quad, \text { for } x=0 \\\\ \frac{\sqrt{x+b x^{2}}-\sqrt{x}}{b x^{\frac{3}{2}}}, \text { for } x>0 \end{array}\right.$

Solution:

$f(x)=\left\{\begin{array}{c} \frac{\sin (a+1) x+\sin x}{x}, \text { for } x<0 \\\\ c \quad, \text { for } x=0 \\\\ \frac{\sqrt{x+b x^{2}}-\sqrt{x}}{b x^{\frac{3}{2}}}, \text { for } x>0 \end{array}\right.$

Since $f\left ( x \right )$ is continuous at $x=0$ , we have

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

(LHL at $x=0$ )

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

\begin{aligned} &=\lim _{x \rightarrow 0} \frac{(\sin a x \times \cos x)}{x}+\lim _{x \rightarrow 0} \frac{(\cos a x \times \sin x)}{x}+\lim _{x \rightarrow 0} \frac{\sin x}{x} \\ &=\lim _{x \rightarrow 0} \frac{\sin a x}{x} \lim _{x \rightarrow 0} \cos x+\lim _{x \rightarrow 0} \cos a x \lim _{x \rightarrow 0} \frac{\sin x}{x}+\lim _{x \rightarrow 0} \frac{\sin x}{x} \end{aligned}

\begin{aligned} &=\lim _{x \rightarrow 0} \frac{\sin a x}{x}(1)+(1) \lim _{x \rightarrow 0} \frac{\sin x}{x}+\lim _{x \rightarrow 0} \frac{\sin x}{x} \\ &=\lim _{x \rightarrow 0} \frac{\sin a x}{a x}(a)+\lim _{x \rightarrow 0} \frac{\sin x}{x}+\lim _{x \rightarrow 0} \frac{\sin x}{x} \end{aligned}

\begin{aligned} &=a \lim _{x \rightarrow 0} \frac{\sin a x}{a x}+\lim _{x \rightarrow 0} \frac{\sin x}{x}+\lim _{x \rightarrow 0} \frac{\sin x}{x} \\ &=a(1)+1+1 \\ &=a+2 \end{aligned}                                                                              $\left[\begin{array}{l} \because \lim _{x \rightarrow 0} \cos x=1 \\ \because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1 \end{array}\right]$

RHL at  $x=0$

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

\begin{aligned} &=\lim _{x \rightarrow 0} \frac{x+b x^{2}-x}{b x \sqrt{x}\left(\sqrt{x+b x^{2}}+\sqrt{x}\right)} \\\\ &=\lim _{x \rightarrow 0} \frac{b x^{2}}{b x \sqrt{x}\left(\sqrt{x+b x^{2}}+\sqrt{x}\right)} \end{aligned}

\begin{aligned} &=\lim _{x \rightarrow 0} \frac{x}{\sqrt{x}\left(\sqrt{x+b x^{2}}+\sqrt{x}\right)} \\\\ &=\lim _{x \rightarrow 0} \frac{\sqrt{x}}{\left(\sqrt{x+b x^{2}}+\sqrt{x}\right)} \end{aligned}

$=\lim _{x \rightarrow 0} \frac{\left(\frac{\sqrt{x}}{\sqrt{x}}\right)}{\left(\frac{\sqrt{x+b x^{2}}+\sqrt{x}}{\sqrt{x}}\right)}$

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

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

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

$a=\frac{-3}{2}, c=\frac{1}{2}, b$  is any real number except  $0$