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STD 12 - 5. continuity and differentiation — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 5. continuity and differentiation4 Marks
Question
The value of $p$ for which the function $f(x) = \left\{ \begin{array}{l}\frac{{{{({4^x} - 1)}^3}}}{{\sin \frac{x}{p}\log \left[ {1 + \frac{{{x^2}}}{3}} \right]}},\,x \ne 0\\\,\,\,\,\,\,\,\,\,\,\,12{(\log 4)^3},\,\,x = 0\end{array} \right.$ may be continuous at $x = 0$, is
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Answer

For $f(x)$ to be continuous at $x = 0,$
we should have $\mathop {\lim }\limits_{x \to 0} \,\,f(x) = f(0) = 12\,{(\log \,4)^3}$
$\mathop {\lim }\limits_{x \to 0} \,\,f(x) = \mathop {\lim }\limits_{x \to 0} \,{\left( {\frac{{{4^x} - 1}}{x}} \right)^3} \times \frac{{\left( {\frac{x}{p}} \right)}}{{\left( {\sin \frac{x}{p}} \right)}}.\frac{{p{x^2}}}{{\log \,\left( {1 + \frac{1}{3}{x^2}} \right)}}$
$ = {(\log 4)^3}\,.\,1\,.\,p\,.\mathop {\lim }\limits_{x \to 0} \,\left( {\frac{{{x^2}}}{{\frac{1}{3}{x^2} - \frac{1}{{18}}{x^4} + .........}}} \right)$
$ = 3p\,\,{(\log 4)^3}.$ Hence $p = 4.$

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