The function $\mathrm{f}$ defined on $\left(-\frac{1}{3}, \frac{1}{3} ight)$ by $\mathrm{f}(x)=\left\{\begin{array}{cc} \frac{1}{x} \log \left(\frac{1+3 x}{1-2 x} ight), & x
eq 0 \ \mathrm{k} & , \quad x=0 \end{array} ight.$ is continuous at $x=0$, then $\mathrm{k}$ is

Question

The function $\mathrm{f}$ defined on $\left(-\frac{1}{3}, \frac{1}{3}ight)$ by $\mathrm{f}(x)=\left\{\begin{array}{cc} \frac{1}{x} \log \left(\frac{1+3 x}{1-2 x}ight), & x 
eq 0 \ \mathrm{k} & , \quad x=0 \end{array}ight.$ is continuous at $x=0$, then $\mathrm{k}$ is

Aicharya · Accepted Answer

$\mathrm{f}$ is continuous at $x=0$. $\begin{aligned} 	herefore \quad \mathrm{f}(0) & =\lim _{x ightarrow 0} \mathrm{f}(x) \ 	herefore \quad \mathrm{k} & =\lim _{x ightarrow 0}\left(\frac{1}{x} \log (1+3 x)-\frac{1}{x} \log (1-2 x)ight) \ & =\lim _{x ightarrow 0}\left(\frac{3 \log (1+3 x)}{3 x}+\frac{2 \log (1-2 x)}{-2 x}ight) \ & =3+2=5 \end{aligned}$

The function $\mathrm{f}$ defined on $\left(-\frac{1}{3}, \frac{1}{3}\right)$ by…

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