MCQ
$\lim _{x \rightarrow 0} \frac{(1+x)^5-1}{(1+x)^3-1}=$
  • A
    $0$
  • B
    1
  • $\frac{5}{3}$
  • D
    $\frac{5}{3}$

Answer

Correct option: C.
$\frac{5}{3}$
(C)
Applying L-Hospital's rule, we get
$\lim _{x \rightarrow 0} \frac{(1+x)^5-1}{(1+x)^3-1}=\lim _{x \rightarrow 0} \frac{5(1+x)^4}{3(1+x)^2}=\frac{5}{3}$

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