Discussion Forum

For what value of k, the function defined by 

$f(x)=\begin{cases}\frac{log(1+2x)\sin x^{0}}{x^{2}} &for x\neq0\\k & for x=0\end{cases}$

 is continuous at x=0?

Answer:

Explanation:

Given , p(x)$=\begin{cases}\frac{log(1+2x)\sin x^{0}}{x^{2}} &for x\neq0\\k & for x=0\end{cases}$

 =  $f(x)=\begin{cases}\frac{log(1+2x)\sin \frac{\pi x}{180}}{x^{2}} &for x\neq0\\k & for x=0\end{cases}$

 Since f(x) is continuous at x=0

 $\therefore$   LHL=f(0)

Now, LHL=  $\lim_{x \rightarrow 0-}f(x)=\lim_{h \rightarrow 0}f(0-h)$

  =$\lim_{h \rightarrow 0}\frac{\log(1+2(0-h)\sin\frac{\pi}{180^{0}}(0-h))}{(0-h)^{2}}$

 =$\lim_{h \rightarrow 0}\frac{\log(1-2h)\left\{-\sin\frac{\pi h}{180}\right\}}{h^{2}}$

 =  $\lim_{h \rightarrow 0}(-2)\frac{\log(1-2h)}{-2h}\times(-)\lim_{h \rightarrow 0}\frac{\sin \frac{\pi h}{180}}{\frac{\pi h}{180}}\times\frac{\pi}{180}$

= $(-2)\times(-1)\times 1\times \frac{\pi}{180}$

$\left[\because \lim_{x \rightarrow 0}\log\frac{1+x}{x}=1 and    \lim_{x \rightarrow 0}\frac{\sin x}{x}=1\right]$

=$\frac{\pi}{90}$