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

Question

Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{2x}-\text{e}^\text{x}}{\sin2\text{x}}$

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Answer

$\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{2x}-\text{e}^\text{x}}{\sin2\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{2x}-\text{1}}{\sin2\text{x}}-\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{x}-\text{1}}{\sin2\text{x}}$ $=\Big(\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{2x}-\text{1}}{2\text{x}}\times\lim\limits_{\text{x}\rightarrow0}\frac{\text{2x}}{\sin2\text{x}}\Big)-\frac{1}{2}\Big(\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{x}-1}{\text{x}}\times\lim\limits_{\text{x}\rightarrow0}\frac{\text{2x}}{\sin2\text{x}}\Big)$ $=1-\frac{1}{2}$ $=\frac12$

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