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According to the definition of $\epsilon$-$\delta$ of a limit we have:

$\forall \epsilon \gt 0$, $\exists \delta \gt 0$ such that

$$0 \lt |x-a| \lt \delta \implies |f(x)-L|\lt \epsilon$$

Now I tried to prove $$\lim_{x \to 0}\frac{\sin x}{x}=1$$

We are given $\epsilon$ and we need to figure out $\delta$

So we have:

$$\left|\frac{\sin x}{x}-1\right|\lt \epsilon$$

Assuming $x \gt 0$ we get

$$|\sin x-x|\lt x\epsilon$$

and for $x \gt 0$ we have $x \gt \sin x$

So

$$x-\sin x\lt x\epsilon$$

$$x(1-\epsilon)\lt \sin x \lt 1$$

So we get

$$x \lt \frac{1}{1-\epsilon}$$

Hence $$\delta=\frac{1}{1-\epsilon}$$

Now I took $\epsilon=0.01$ to get $\delta=\frac{100}{99}=1.0101\cdots$

Now $$\left|\frac{\sin 1.0101...}{1.0101...}-1\right|=|0.8384-1|=0.161$$

But $$0.161 \gt \epsilon=0.01$$

Why is this contradictory?

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There is no reason for your $\delta$ to works. Your are after a $\delta>0$ such that$$\lvert x\rvert<\delta\implies\left\lvert\frac{\sin(x)}x-1\right\rvert<\varepsilon.$$You concluded from the inequality$$\left\lvert\frac{\sin(x)}x-1\right\rvert<\varepsilon.\tag1$$that, if $x>0$, then $x<\frac1{1-\varepsilon}$. But you are not supposed to extract conclusions from $(1)$. Instead, you are are supposed to prove that it holds if $\lvert x\rvert<\delta$, for some $\delta>0$.

I must say that I don't think that it is a good idea to prove that $\lim_{x\to0}\frac{\sin(x)}x=1$ using the $\varepsilon-\delta$ definition.

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  • $\begingroup$ ok can you recommend any good book which explains this concept of $\epsilon-\delta$ with lot of examples $\endgroup$ – Umesh shankar Jun 28 '19 at 18:48
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    $\begingroup$ @Umesh shankar, use the book "Examples and Theorems in Analysis" by Peter Walker or "Analysis I" by Terence Tao. $\endgroup$ – Pratik Apshinge Jun 28 '19 at 19:37
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You found that $$ x(1 - \epsilon) < \sin x ,$$ which implies that $$ x < \frac {\sin x}{1-\epsilon}.$$

The way you constructed this, it is a tight bound, meaning there is no room for any additional error. If you choose any number greater than $ \frac {\sin x}{1-\epsilon},$ you will end up outside the epsilon bounds that you set.

And then you went and chose a number that you knew was greater than $ \frac {\sin x}{1-\epsilon}. $ So it is not surprising at all that it did not work.

Here is a hint for making problems like this easier. If you overestimate $\delta$ it will give you a wrong result every time. But there is no penalty (at least none mathematically) for underestimating $\delta.$ If the exact bound is hard to calculate, as it is here, go for something between zero and the exact bound.

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