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Limits — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsLimits3 Marks
Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}^2\big(1-\cos\text{x}^2\big)}{\text{x}^6}$
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Answer

$\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}^2\big(1-\cos\text{x}^2\big)}{\text{x}^6}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}^2\times2\sin^2\frac{\text{x}^2}{2}}{\text{x}^6}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}^2}{\text{x}^2}\times\lim\limits_{\text{x}\rightarrow0}\frac{2\sin^2\frac{\text{x}^2}{2}}{\text{x}^4}$
$=\Big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}^2}{\text{x}}\Big)\times2\times\Bigg(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\frac{\text{x}^2}{2}}{\frac{\text{x}^2}{2}}\Bigg)\times\frac14$
$=(1)^2\times2\times1\times\frac14$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\Big]$
$=\frac12$

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