MCQ
If $f(x) = \frac{{\sin ({e^{x - 2}} - 1)}}{{\log (x - 1)}},$ then $\mathop {\lim }\limits_{x \to 2} f(x)$ is given by
- A$-2$
- B$-1$
- C$0$
- ✓$1$
$ = \mathop {\lim }\limits_{t \to 0} \,\,\frac{{\sin \,({e^t} - 1)}}{{\log \,(1 + t)}}$, $\{$Putting $x = 2 + t\} $
$ = \mathop {\lim }\limits_{t \to 0} \,\,\frac{{\sin \,({e^t} - 1)}}{{{e^t} - 1}}.\frac{{{e^t} - 1}}{t}.\frac{t}{{\log \,(1 + t)}}$
$ = \mathop {\lim }\limits_{t \to 0} \,\,\frac{{\sin \,({e^t} - 1)}}{{{e^t} - 1}}.\left( {\frac{1}{{1\,\,!}} + \frac{t}{{2\,\,!}} + ...} \right) \times \left[ {\frac{1}{{\left( {1 - \frac{1}{2}t + \frac{1}{3}{t^2} - ...} \right)}}} \right]$
$ = 1\,\,.\,\,1\,\,.\,\,1 = 1,\,\,\,\,(\because \,\,{\text{As}}\,\,t \to 0,\,\,{e^t} - 1 \to 0).$
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$D^*f(x) =\mathop {Limit}\limits_{h \to 0} \frac{{{f^2}(x + h) - {f^2}(x)}}{h}$ where $f^2(x)$ means $[f(x)]^2.$ If $f(x) = x lnx$ then
${\left. {D^*f(x)} \right|_{x = e}}$ has the value
$(\alpha + p)^{m - 1} + (\alpha + p)^{m - 2} (\alpha + q) + (\alpha + p)^{m - 3} (\alpha + q)^2 + ...... (\alpha + q)^{m - 1}$
where $\alpha \ne - q$ and $p \ne q$ is :