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Limits — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsLimits3 Marks
Question
Evaluate the following limits: $\lim _{x \rightarrow 0}\left[\frac{\left(2^x-1\right)^3}{\left(3^x-1\right) \cdot \sin x \cdot \log (1+x)}\right]$
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Answer

$\lim _{x \rightarrow 0} \frac{\left(2^x-1\right)^3}{\left(3^x-1\right) \cdot \sin x \cdot \log (1+x)}$
$=\lim _{x \rightarrow 0} \frac{\frac{\left(2^x-1\right)^3}{x^3}}{\frac{\left(3^x-1\right) \cdot \sin x \cdot \log (1+x)}{x^3}}$
${\left[\begin{array}{l} \text { Divide numerator and } \\ \text { denominator by } x^3 . \\ \because x \rightarrow 0, x \neq 0 \\ \therefore x^3 \neq 0 \end{array}\right]}$
$=\frac{\lim _{x \rightarrow 0}\left(\frac{2^x-1}{x}\right)^3}{\lim _{x \rightarrow 0}\left(\frac{3^x-1}{x}\right) \cdot \frac{\sin x}{x} \cdot \frac{\log (1+x)}{x}}$
$=\frac{\left(\lim _{x \rightarrow 0} \frac{2^x-1}{x}\right)^3}{\left(\lim _{x \rightarrow 0} \frac{3^x-1}{x}\right) \cdot\left(\lim _{x \rightarrow 0} \frac{\sin x}{x}\right)\left(\lim _{x \rightarrow 0} \frac{\log (1+x)}{x}\right)}$
$=\frac{(\log 2)^3}{(\log 3)(1)(1)}$
$\ldots\left[\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1, \lim _{x \rightarrow 0} \frac{\mathbf{a}^x-1}{x}=\log a, \lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1\right]$
$=\frac{(\log 2)^3}{\log 3}$

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