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Limits — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSLimits4 Marks
Question
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{1-\cos2\text{x}+\tan^2\text{x}}{\text{x}\sin\text{x}}$
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Answer

$\lim\limits_{\text{x}\rightarrow0}\frac{1-\cos2\text{x}+\tan^2\text{x}}{\text{x}\sin\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin^2\text{x}+\tan^2\text{x}}{\text{x}\sin\text{x}}$ $=\frac{2\lim\limits_{\text{x}\rightarrow0}\sin^2\text{x}+\lim\limits_{\text{x}\rightarrow0}\tan^2\text{x}}{\lim\limits_{\text{x}\rightarrow0}\text{x}\sin\text{x}}$ $=\frac{\Big(2\big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}\big)^2\times\text{x}^2\Big)+\big(\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}\big)^2\times\text{x}^2}{\big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}\big)\times\text{x}^2}$ $=\frac{\big(2\times1\times\text{x}^2\big)+\big(1\times\text{x}^2\big)}{\big(1\times\text{x}^2\big)}$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\text{ and }\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}=1\Big]$ $=\frac{3\text{x}^2}{\text{x}^2}$ $=3$

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