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Limits — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsLimits1 Mark
MCQ
$\lim _{x \rightarrow 0}\left[\frac{\left(3^{\sin x}-1\right)^3}{\left(3^x-1\right) \cdot \tan x \cdot \log (1+x)}\right]=$
  • A
    $3 \log 3$
  • B
    $2 \log 3$
  • ✓
    $(\log 3)^2$
  • D
    $(\log 3)^3$
✓

Answer

Correct option: C.
$(\log 3)^2$
(C) $(\log 3)^2$
Hint:
$\lim _{x \rightarrow 0} \frac{\left(3^{\operatorname{lin} x}-1\right)^3}{\left(3^x-1\right) \cdot \tan x \cdot \log (1+x)}$
$=\frac{\lim _{x \rightarrow 0} \frac{\left(3^{\sin x}-1\right)^3}{\sin ^3 x} \cdot \frac{\sin ^3 x}{x^3}}{\lim _{x \rightarrow 0}\left(\frac{3^x-1}{x}\right)\left(\frac{\tan x}{x}\right) \cdot \frac{\log (1+x)}{x}} $
$=\frac{\lim _{x \rightarrow 0}\left(\frac{3^{\operatorname{tin} x}-1}{\sin x}\right)^3 \cdot \lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^3}{\lim _{x \rightarrow 0}\left(\frac{3^x-1}{x}\right) \cdot \lim \left(\frac{\tan x}{x}\right) \cdot \lim _{x \rightarrow 0} \frac{\log (1+x)}{x}} $
$=(\log 3)^2$

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