MCQ
$\mathop {\lim }\limits_{x \to \infty } {\left( {1 - \frac{4}{{x - 1}}} \right)^{3x - 1}} = $
  • A
    ${e^{12}}$
  • ${e^{ - 12}}$
  • C
    ${e^4}$
  • D
    ${e^3}$

Answer

Correct option: B.
${e^{ - 12}}$
b
(b) $\mathop {\lim }\limits_{x \to \infty } {\left( {1 - \frac{4}{{x - 1}}} \right)^{3x - 1}} = \mathop {\lim }\limits_{x \to \infty } {\left[ {{{\left( {1 + \frac{{( - 4)}}{{x - 1}}} \right)}^{\left( {\frac{{x - 1}}{{ - 4}}} \right)}}} \right]^{\left( {\frac{{ - 4}}{{x - 1}}} \right)(3x - 1)}}$

$ = {e^{\mathop {\lim }\limits_{x \to \infty } \left[ {\frac{{ - 4\left( {3 - \frac{1}{x}} \right)}}{{\left( {1 - \frac{1}{x}} \right)}}} \right]}} = {e^{ - 12}}$.

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