If the function f (x) defined by
$f(x)=\left\{\begin{array}{cl}\frac{\log (1+3 x)-\log (1-2 x)}{x}, & x \neq 0 \\ k & , x=0\end{array}\right.$ is continuous at $x=0$, then $k=$
(a) 1
(b) 5
(c) $-1$
(d) none of these
(b) 5
Given: $f(x)=\left\{\begin{array}{l}\frac{\log (1+3 x)-\log (1-2 x)}{x}, x \neq 0 \\ k, x=0\end{array}\right.$
If $f(x)$ is continuous at $x=0$, then $\lim _{x \rightarrow 0} f(x)=f(0)$.
$\Rightarrow \lim _{x \rightarrow 0}\left(\frac{\log (1+3 x)-\log (1-2 x)}{x}\right)=k$
$\Rightarrow \lim _{x \rightarrow 0}\left(\frac{3 \log (1+3 x)}{3 x}-\frac{2 \log (1-2 x)}{2 x}\right)=k$
$\Rightarrow 3 \lim _{x \rightarrow 0}\left(\frac{\log (1+3 x)}{3 x}\right)-2 \lim _{x \rightarrow 0}\left(\frac{\log (1-2 x)}{2 x}\right)=k$
$\Rightarrow 3 \lim _{x \rightarrow 0}\left(\frac{\log (1+3 x)}{3 x}\right)+2 \lim _{x \rightarrow 0}\left(\frac{\log (1-2 x)}{-2 x}\right)=k$
$\Rightarrow 3 \times 1+2 \times 1=k$ $\left[\because \lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1\right]$
$\Rightarrow k=3+2=5$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.