# Epsilon delta proof for this function

I am having a lot of trouble finding an epsilon delta proof for the following: $$\lim_{t\rightarrow0} \hspace{5px} \dfrac{\sin(t^2)}{2t^2} = \dfrac{1}{2}$$

I have tried a lot of things and I can show that $$|\dfrac{\sin(t^2)}{2t^2} - \dfrac{1}{2}| \leq |\dfrac{1}{t^2}| + \dfrac{1}{2}$$ but I am unsure how to go from here to find $$\delta$$ as a function of $$\epsilon$$.

• Can you use the fact that $\lim_{t\to0}\frac{\sin t}t=1$? – José Carlos Santos May 11 '19 at 11:32
• Is it essential to use this fact in the proof? – user208480 May 11 '19 at 11:38
• No, but which properties of the sine function can you use? – José Carlos Santos May 11 '19 at 11:39
• I am guessing using that limit is fine but if there is a simple epsilon delta proof without using it would help to see. – user208480 May 11 '19 at 11:41
• @user208480: to solve this exercise, there needs to be some definition of the sine function. Telling that the limit $\sin t/t=1$ is a minimalistic way to define it (or at least give a useful property). What is the definition in your course ? – Yves Daoust May 11 '19 at 11:59

Take $$\varepsilon>0$$. Since $$\lim_{t\to0}\dfrac{\sin t}t=1$$, there is soma $$\delta'>0$$ such that $$\lvert t\rvert<\delta'\implies\left\lvert\dfrac{\sin t}t-1\right\rvert<2\varepsilon$$. Let $$\delta=\sqrt{\delta'}$$. Then$$\lvert t\rvert<\delta\implies\lvert t^2\rvert<\delta'\implies\left\lvert\frac{\sin(t^2)}{2t^2}-\frac12\right\rvert<\varepsilon.$$
When you need an $$\epsilon, \delta$$ proof you should always look for relevant inequalities. In this case the following inequality $$\sin x is your friend. Combined with this and the fact that $$\cos x =1-2\sin^2(x/2)$$ we get the following inequality $$\cos x>1-\frac{x^2}{2},0 Let an $$\epsilon>0$$ be given and our goal is ensure that the following inequality $$\left|\frac{\sin t^2}{2t^2}-\frac{1}{2}\right|<\epsilon$$ or equivalently $$\left|\frac{\sin t^2}{t^2}-1\right|<2\epsilon$$ holds by constraining the values of $$t$$ under an inequality $$0<|t|<\delta$$ for some suitable $$\delta>0$$.
Let's put $$z=t^2$$ to simplify typing and then our job is to find a $$\delta>0$$ such that the following implication holds $$0 If $$\delta<\sqrt{\pi/2}$$ and $$0 then we have via $$(1)$$ $$\cos z<\frac{\sin z} {z} <1$$ and then using $$(2)$$ we get $$1-\frac{z^2}{2}<\frac{\sin z} {z} <1$$ and then $$\left|\frac{\sin z} {z} - 1\right|=1-\frac{\sin z} {z} <\frac{z^2}{2}$$ If $$z^2<4\epsilon$$ then our job is done. It follows that we can take $$\delta=\min(\sqrt {\pi/2},2\sqrt{\epsilon})$$ and then we have the implication $$0<|t|<\delta\implies\left|\frac{\sin t^2}{2t^2}-\frac{1}{2}\right|<\epsilon$$