Question
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}^3\cot\text{x}}{1-\cos\text{x}}$

Answer

$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}^3\cot\text{x}}{1-\cos\text{x}}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{\text{x}^3}{\tan\text{x}(1-\cos\text{x})}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{\text{x}^3}{\tan\text{x}.2\sin^2\frac{\text{x}}{2}}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{1}{\frac{\tan\text{x}}{\text{x}}\times\frac{2\sin^2\frac{\text{x}}{2}}{\text{x}^2}}$ $=\frac{1}{\bigg(\lim\limits_{\text{x} \rightarrow0}\frac{\tan\text{x}}{\text{x}}\bigg)\times2\bigg(\lim\limits_{\text{x} \rightarrow0}\frac{\sin\frac{\text{x}}{2}}{\frac{\text{x}}{2}}\bigg)^2\times\frac14}$ $=\frac{1}{2\times2\times1\times\frac14}$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1,\lim\limits_{\theta\rightarrow0}\frac{\tan\theta}{\theta}=1\Big]$ $=2$

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