#### Explain solution RD Sharma class 12 chapter Continuity exercise 8.2 question 3 subquestion (ii) maths

x = 2

Hint:

To check the continuity of such type of function we check at the breaking point.

Given:

$f(x)= \begin{cases}\frac{x^{4}-16}{x-2} & x \neq 2 \\ 16 & x=2\end{cases}$

Explanation:

At point x =2

\begin{aligned} &\text { L.H.L }=\lim _{x \rightarrow 2^{-}} f(x)=\lim _{h \rightarrow 0}(2-h)=\lim _{h \rightarrow 0} \frac{(2-h)^{4}-16}{(2-h)-2} \\ &=\lim _{h \rightarrow 0} \frac{2^{4}-4.8 h+6.4 h^{2}-4.2 h^{3}+h^{4}-16}{-h} \end{aligned}

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{32-32 h+24 h^{2}-8 h^{3}+h^{4}-16}{-h} \\ &=\lim _{h \rightarrow 0} 32-24 h+8 h^{2}-h^{3}=32 \end{aligned}

$\text { R.H.L }=\lim _{x \rightarrow 2^{+}} f(x)=\lim _{h \rightarrow 0}(2+h)=\lim _{h \rightarrow 0} \frac{(2+h)^{4}-16}{(2+h)-2}$

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{2^{4}+4.8 h+6.4 h^{2}+4.2 h^{3}+h^{4}-16}{h}=32 \\ &\text { Also } f(2)=16 \\ &\text { Thus } L . H .L=R . H . L \neq f(2) \end{aligned}

Therefore, f(x) is discontinuous at x = 2