Evaluate the following limit: _ x 0 2x- 2x x^3 - Vidyadip

Question

Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\tan2\text{x}-\sin2\text{x}}{\text{x}^3}$

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Answer

$\lim\limits_{\text{x}\rightarrow0}\frac{\tan2\text{x}-\sin2\text{x}}{\text{x}^3}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\frac{\sin2\text{x}}{\cos2\text{x}}-\sin2\text{x}}{\text{x}^3}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\sin2\text{x}\big(\frac{1}{\cos2\text{x}}-1\big)}{\text{x}^3}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\sin2\text{x}(1-\cos2\text{x})}{\text{x}^2\cos2\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\sin2\text{x}\big(2\sin^2\text{x}\big)}{\text{x}^3\cos2\text{x}}$ $=\frac{\big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin2\text{x}}{\text{x}}\big)\Big(\lim\limits_{\text{x}\rightarrow0}\frac{2\sin^2\text{x}}{\text{x}^2}\Big)}{\big(\lim\limits_{\text{x}\rightarrow0}\cos2\text{x}\big)}$ $=\frac{\big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin2\text{x}}{\text{x}}\times2\big)\Big(\lim\limits_{\text{x}\rightarrow0}2\big(\frac{\sin\text{x}}{\text{x}}\big)^2\Big)}{\lim\limits_{\text{x}\rightarrow0}\cos2\text{x}}$ $=\frac{(2\times1)(2\times1)}{1}$ $=4$

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