MCQ
જો $f(x) = \frac{{\sin ({e^{x - 2}} - 1)}}{{\log (x - 1)}}$ તો $\mathop {\lim }\limits_{x \to 2} f(x) = . . . $
- 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).$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.