Question

find the value of k, if the function are continuous at the point indicates $\begin{array}{c}f\left(x\right)={\log }_{\left(1-2x\right)}\left(1+2x\right)\\ =k\end{array}$
for $\begin{array}{c}x\ne 0\\ x=0\end{array}$

Answer 1

$f\left(x\right)=log(1+2x)$

$(1-2x)$ , $x\ne o$

$=k$ , $x=0$

To be continueous at $x=0,sinf\left(x\right)=f\left(o\right).$

$x\to o$

$Nowsinf\left(x\right)=sinlog(1+2x)(1-2x)$

$x\to o$ $x\to o$

$=sin\frac{log(1+2x)}{log(1-2x)}$ $|\frac{o}{o}form$

$x\to o$

$\underset{x\to o}{sin}\frac{\frac{1}{1+2x}.2}{\frac{1}{1-2x}-2}|L^{\prime }Hospita{l}^{\prime }sRule$

$=\frac{2}{-2}$

$=-1$

$\therefore sinf\left(x\right)=f\left(o\right).$

$x\to o$

$x\Rightarrow -1=k$

$\therefore k=-1$