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STD 11 - 12. limits — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 12. limits4 Marks
Question
If $f(x) = \frac{{\sin ({e^{x - 2}} - 1)}}{{\log (x - 1)}},$ then $\mathop {\lim }\limits_{x \to 2} f(x)$ is given by
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Answer

d
(d) $\mathop {\lim }\limits_{x \to 2} \,\,f(x) = \mathop {\lim }\limits_{x \to 2} \,\frac{{\sin \,({e^{x - 2}} - 1)}}{{\log \,(x - 1)}}$

$ = \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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