# Showing $\lim_{x \to 0}\frac{\sin x}{x}=1$ with epsilon-delta [duplicate]

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 $$0.161 \gt \epsilon=0.01$$

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.

• ok can you recommend any good book which explains this concept of $\epsilon-\delta$ with lot of examples – Umesh shankar Jun 28 '19 at 18:48
• @Umesh shankar, use the book "Examples and Theorems in Analysis" by Peter Walker or "Analysis I" by Terence Tao. – Pratik Apshinge Jun 28 '19 at 19:37

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.