$$\lim_{x\to0}{\frac{sinx}{x}}=1$$ or can only use some other methods like the l'Hôpital's Rule to solve it
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1Using l'Hopital on this limit is dangerous. The reason is that you need to know the derivative of $\sin x$, and in the process of finding the derivative of $\sin x$ by evaluating $\lim_{h\to 0}\frac{\sin(x+h) - \sin x}{h}$, you will have to find the limit $\lim_{h\to 0}\frac{\sin h}{h}$. There are better ways, though. – Arthur Oct 16 '17 at 11:02
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1Watch this. – Faiq Irfan Oct 16 '17 at 11:02
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Sandwitch Theorem, of course. – Faiq Irfan Oct 16 '17 at 11:02
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1Possible duplicate of How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$? – Guy Fsone Oct 16 '17 at 11:22
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I prove that the limit is $1$ using $\varepsilon$-$\delta$ definition.
Sketch Work:
Since $$\cos x<\frac{\sin x}{x}<1,$$ we have $$\bigg|\frac{\sin x}{x}-1\bigg|<1-\cos x$$ and $1-\cos x=2\sin^2\frac{x}{2}\leqslant\frac{x^2}{2}$ hence $$\bigg|\frac{\sin x}{x}-1\bigg|\leqslant\frac{x^2}{2}$$
Solution:
Let $\varepsilon>0$ be given. Choose $\delta = \sqrt{2\varepsilon}>0.$ Then for any $0<|x|<\delta = \sqrt{2\varepsilon},$ we have $$x^2<2\varepsilon.$$ Therefore, $$\bigg| \frac{\sin x}{x} - 1 \bigg| \leq \frac{x^2}{2}<\varepsilon.$$ Hence, $$\lim_{x\to 0}\frac{\sin x}{x} = 1.$$
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