If the function $f$ defined on $\left(-\frac{1}{3}, \frac{1}{3}\right)$ by $f(x)=\left\{\begin{array

Question

If the function $f$ defined on $\left(-\frac{1}{3}, \frac{1}{3}ight)$ by $f(x)=\left\{\begin{array}{cc}\frac{1}{x} \log _{e}\left(\frac{1+3 x}{1-2 x}ight), & 	ext { when } x 
eq 0 \ k, & 	ext { when } x=0\end{array}ight.$ is continuous, then $k$ is equal to

Professor PC Thomas Classes · Answer

(5) :\ $\lim _{x ightarrow 0} f(x)=\lim _{x ightarrow 0}\left(\frac{1}{x} \log _{e}\left(\frac{1+3 x}{1-2 x}ight)ight)$\ $=\lim _{x ightarrow 0}\left(\frac{\log _{e}(1+3 x)}{x}-\frac{\log _{e}(1-2 x)}{x}ight)$\ $=\lim _{x ightarrow 0}\left(\frac{3 \log _{e}(1+3 x)}{3 x}+\frac{2 \log _{e}(1-2 x)}{-2 x}ight)$\ $=3+2=5$\ Since $f(x)$ is continuous.\ $	herefore f(0)=\lim _{x ightarrow 0} f(x) \Rightarrow k=5$\