# $\lim_{x \rightarrow 0} \frac{\sin(x)}{x}$ using $\epsilon - \delta$ definition

So I'm trying to evaluate this limit without using squeeze theorem and doing it by $\epsilon - \delta$ definition.

Here's my attempt:

$$\left| \frac{\sin(x)}{x} - 1 \right| \leq \left| \frac{\sin(x)}{x}\right| + |1| \leq \frac{1}{|x|} + 1$$

Let $\epsilon > 0$ and $\delta = \frac{1}{\epsilon}$

$$|x| < \delta \Rightarrow \frac{1}{|x|} + 1 < \epsilon + 1$$

And I'm stuck here. I also tried $\delta = \frac{1}{\epsilon - 1}$ but that won't work since I'll only proof for $\epsilon > 1$.

Can someone help me? Thanks.

• What are you allowed to use? Are you allowed the Taylor expansion of $\sin$ at $0$? Are you allowed that $\sin(0) = 0$ and that $\frac{\mathrm{d}}{\mathrm{d}x} \sin(x) |_{x = 0} = 1$ and that $\sin$ is differentiable? Commented Apr 12, 2016 at 13:37
• Your technique is bound to fail. The only thing you're using about sin(x) is that |sin(x)/x)| is less than or equal to 1/x, but 1/x blows up at x = 0 so this bound can't give you that the limit exists (let alone help you compute it). In other words, no algebra trick is going to work if the only thing you use about sin is that it's less than 1, you have to use something deeper about the sin function. Commented Apr 12, 2016 at 13:37
• See the answers in math.stackexchange.com/questions/285313/… Commented Apr 12, 2016 at 14:01

Since $\cos\theta<\frac{\sin\theta}{\theta}<1$ $$\bigg|\frac{\sin\theta}{\theta}-1\bigg|<1-\cos\theta$$ and $1-\cos\theta=2\sin^2\frac{\theta}{2}\leqslant\frac{\theta^2}{2}$ hence $$\bigg|\frac{\sin\theta}{\theta}-1\bigg|\leqslant\frac{\theta^2}{2}$$
• How can I proof the first inequality? I verified numerically that it is valid for $| \theta | < 4.4934...$ but I failed to proof this. Despite that I believe I did the exercise, choosing $\delta = \sqrt{2 \epsilon}$.