MCQ
$\mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + x)}^5} - 1}}{{{{(1 + x)}^3} - 1}} = $
  • A
    $0$
  • B
    $1$
  • $5/3$
  • D
    $3/5$

Answer

Correct option: C.
$5/3$
c
(c) $\mathop {\lim }\limits_{x \to 0} \,\frac{{x\,{[^5}{C_1}{ + ^5}{C_2}x{ + ^5}{C_3}{x^2}{ + ^5}{C_4}{x^3}{ + ^5}{C_5}{x^4}]}}{{x\,{[^3}{C_1}{ + ^3}{C_2}x{ + ^3}{C_3}{x^2}]}}$ $ = \frac{5}{3}.$

Aliter : Apply  $ L$-Hospital’s rule.

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