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Continuity and Differentiability — MATHS STD 12 Science — Question

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

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

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