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${e^5}$
✓
Answer
Correct option: C.
${e^{ - 5}}$
c
(c) $\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{x + 2 - 5}}{{x + 2}}} \right)^x} = \mathop {\lim }\limits_{x \to \infty } {\left[ {{{\left( {1 - \frac{5}{{x + 2}}} \right)}^{\frac{{x + 2}}{{ - 5}}}}} \right]^{ - \,\frac{{5x}}{{x + 2}}}} = {e^{ - 5}}$
(c) $\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{x + 2 - 5}}{{x + 2}}} \right)^x} = \mathop {\lim }\limits_{x \to \infty } {\left[ {{{\left( {1 - \frac{5}{{x + 2}}} \right)}^{\frac{{x + 2}}{{ - 5}}}}} \right]^{ - \,\frac{{5x}}{{x + 2}}}} = {e^{ - 5}}$
$\left( {\because \,\mathop {\lim }\limits_{x \to \infty } \frac{{ - 5x}}{{x + 2}} = \,\mathop {\lim }\limits_{x \to \infty } \frac{{ - 5}}{{1 + \frac{2}{x}}} = - 5} \right)$.
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