text: #### Please solve RD Sharma class 12 chapter Continuity exercise 8.2 question 3 subquestion (iii) maths textbook solution
  
  x = 0
  
  Hint:
  
  $\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$
  
  Given:
  
  $f(x)= \begin{cases}\frac{\sin x}{x} & x<0 \\ 2 x+3 & x \geq 0\end{cases}$
  
  Explanation:
  
  At x = 0
  
  \begin{aligned} &\text { L.H.L }=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0}(0-h)=\lim _{h \rightarrow 0} \frac{\sin (-h)}{-h} \\ &=\lim _{h \rightarrow 0} \frac{-\sinh }{-h}=1 \end{aligned}
  
  \begin{aligned} &\text { R.H.L }=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0}(0+h)=\lim _{h \rightarrow 0} \frac{\sinh }{h}=1 \\ &\text { Also } f(0)=2 \times 0+3=3 \\ &\text { Thus L.H.L}=\text {R.H.L} \neq \mathrm{f}(0) \end{aligned}
  
  f is discontinuous at x = 0