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Question Number 121050 by mathocean1 last updated on 05/Nov/20

$$\mathrm{show}\mathrm{that}$$$$\underset{\mathrm{x}\to 0}{\lim }\frac{\mathrm{sinx}}{\mathrm{x}}=1$$

Answered by 675480065 last updated on 05/Nov/20

$$\mathrm{as}\mathrm{x}\to 0,\mathrm{sinx}\to (\mathrm{x}-\frac{{\mathrm{x}}^3}{3!}+0\left({\mathrm{x}}^3\right))$$$$\Rightarrow \underset{x\to 0}{\lim }\frac{\mathrm{sinx}}{\mathrm{x}}=\underset{x\to 0}{\lim }\frac{\mathrm{x}(1-\frac{{\mathrm{x}}^2}{6})}{\mathrm{x}}=\underset{x\to 0}{\lim }(1-\frac{{\mathrm{x}}^2}{6})=1$$

Answered by physicstutes last updated on 05/Nov/20

$$\mathrm{as}x\to 0,\sin x\to x$$$$\Rightarrow \underset{x\to 0}{\lim }\frac{x}{x}=1\mathrm{QED}$$

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