MCQ
For $x \in R,\,\,\,\mathop {\lim }\limits_{x \to \infty } \,{\left( {\frac{{x - 3}}{{x + 2}}} \right)^x}$ is equal to
- A$e$
- B${e^{ - 1}}$
- ✓${e^{ - 5}}$
- D${e^5}$
$\left( {\because \,\mathop {\lim }\limits_{x \to \infty } \frac{{ - 5x}}{{x + 2}} = \,\mathop {\lim }\limits_{x \to \infty } \frac{{ - 5}}{{1 + \frac{2}{x}}} = - 5} \right)$.
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$f(x)=\sin \left(\frac{\pi x}{12}\right) \text { and } g(x)=\frac{2 \log _{ e }(\sqrt{x}-\sqrt{\alpha})}{\log _{ e }\left( e ^{\sqrt{x}}- e ^{\sqrt{\alpha}}\right)} \text {. }$
Then the value of $\lim _{ x \rightarrow \alpha^{+}} f( g ( x ))$ is