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Limits — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsLimits2 Marks
Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{3\sin2\text{x}+2\text{x}}{3\text{x}+2\tan3\text{x}}$
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Answer

$\lim\limits_{\text{x}\rightarrow0}\frac{3\sin2\text{x}+2\text{x}}{3\text{x}+2\tan3\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{3\frac{\sin\text{x}}{\text{x}}+2}{3+2\tan\frac{3\text{x}}{\text{x}}}$
$=\frac{\lim\limits_{\text{x}\rightarrow0}\frac{3\sin2\text{x}}{\text{x}}+\lim\limits_{\text{x}\rightarrow0}2}{\lim\limits_{\text{x}\rightarrow0}3+\lim\limits_{\text{x}\rightarrow0}\frac{2\tan3\text{x}}{\text{x}}}$
$=\frac{\Big(3\lim\limits_{2\text{x}\rightarrow0}\frac{\sin2\text{x}}{2\text{x}}\times2\Big)+2}{3+\Big(2\times\lim\limits_{\text{x}\rightarrow0}\frac{\tan3\text{x}}{3\text{x}}\times3\Big)}$
$=\frac{(3\times2)+2}{3+(2\times3)}$ $\Big[\because\ \lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\text{ also }\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}=1\Big]$
$=\frac{8}{9}$

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