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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability4 Marks
Question
If given function is continuous at $x =0$ then write the value of $k. f(x)=\left\{\begin{array}{cc} \frac{\log (1+a x)-\log (1-b x)}{x}, & x \neq 0 \\ k, & x=0 \end{array}\right.$
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Answer

value of $\text{R.H.L}$. at $x=0$
$\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} \frac{\log (1+a(0+h))-\log (1-b(0+h))}{(0+h)}$
$=\lim _{h \rightarrow 0} \frac{\log (1+a h)-\log (1-b h)}{h}$
$= \lim _{h \rightarrow 0} \frac{\left(a h-\frac{(a h)^2}{2}+\frac{(a h)^3}{3}-\ldots\right)-\left(-b h-\frac{(b h)^2}{2}-\frac{(b h)^3}{3}-\ldots\right)}{h}$
$= \lim _{h \rightarrow 0}\left(\frac{(a+b) h-\frac{h^2}{2}\left(a^2-b^2\right)+\frac{h^3}{3}\left(a^3+b^3\right)-\ldots \ldots .}{h}\right]$
$= \lim _{h \rightarrow 0}(a+b) h\left[\frac{1-\frac{h}{2}(a-b)+\frac{h^2}{3}\left(a^2-a b+b^2\right)-\ldots \ldots .}{h}\right]$
$=a+b$
and $ f(0)=k$
$\because$ function is continuous at $x=0$
$\therefore f(0)=\lim _{h \rightarrow 0} f(0+h)$
$\therefore k=a+b \text {} $

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