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 $\epsilon$.
On
When you need an $\epsilon, \delta$ proof you should always look for relevant inequalities. In this case the following inequality $$\sin x <x<\tan x, 0<x<\frac{\pi}{2}\tag{1}$$ is your friend. Combined with this and the fact that $\cos x =1-2\sin^2(x/2)$ we get the following inequality $$\cos x>1-\frac{x^2}{2},0<x<\frac{\pi}{2}\tag{2}$$ 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 $$0<z<\delta^2\implies\left|\frac{\sin z} {z} - 1\right|<2\epsilon$$ If $\delta<\sqrt{\pi/2}$ and $0<z<\delta^2<\pi/2$ 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 $$0<|t|<\delta\implies\left|\frac{\sin t^2}{2t^2}-\frac{1}{2}\right|<\epsilon$$
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$$\lvert t\rvert<\delta\implies\lvert t^2\rvert<\delta'\implies\left\lvert\frac{\sin(t^2)}{2t^2}-\frac12\right\rvert<\varepsilon.$$