$$\lim_{x\to 0} \frac{1-\cos(x)}{x^2}=\frac{1}{2}$$
Proof: $$\lim_{x\to 0} \frac{1-\cos (x)}{x^2} \times \frac{1+\cos (x)}{1+\cos(x)}$$ $$\lim_{x\to 0} \frac{1-\cos^2(x)}{x^2(1+\cos (x))}$$ $$\lim_{x\to 0} \frac{\sin^2(x)}{x^2(1+\cos (x))}$$ $$=\frac{1}{2}$$ Hence we prove it. But I try to use $\varepsilon$ & $\delta$ method but I was not able to complete the proof. Some one help me to prove it using $\varepsilon$ and $\delta$ method. Thanks in advance.
Proving any and every limit via $\epsilon-\delta$ definition is not a good idea and it does not help at all in understanding the $\epsilon-\delta$ definition of limit. Still many textbooks and teachers give problems to prove a limit by $\epsilon-\delta$. And students most of the time struggle to find an expression for $\delta$ in terms of $\epsilon$. This kind of approach merely treats calculus/analysis as a fancy extension of algebra where you have to do all sorts of algebraic symbol shunting in order to get $\delta$ in terms of $\epsilon$. This is so wrong an approach. An understanding of calculus can not be had by treating it in terms of algebraic manipulation. You have to transcend the algebraic thought process and really start having an appreciation for the concept of real numbers and order relations ($<, >$) on real numbers.
Further the right way to understand the $\epsilon-\delta$ definition of limit is to use it to verify simple limits like $\lim_{x \to 2}x^{2} = 4$ and more importantly to see the use of this definition in proving limit theorems. Moreover the emphasis should not be on finding a $\delta$ as an explicit expression in $\epsilon$ but rather one should be convinced that for any $\epsilon$ a corresponding $\delta$ exists without ever writing $\delta$ explicitly in terms of $\epsilon$.
But still you insist to have an $\epsilon-\delta$ proof for the limit $$\lim_{x \to 0}\frac{1 - \cos x}{x^{2}} = \frac{1}{2}\tag{1}$$ then the only way out is to use the standard limits $$\lim_{x \to 0}\cos x = \lim_{x \to 0}\frac{\sin x}{x} = 1\tag{2}$$ and to use the algebraic identity $$\frac{1 - \cos x}{x^{2}} = \dfrac{\dfrac{\sin x}{x}\cdot\dfrac{\sin x}{x}}{1 + \cos x}\tag{3}$$ which you have already established in your question.
In case you have studied the proof of limit theorems concerning algebra of limits then you can see that via equation $(3)$ the desired limit $(1)$ can be evaluated by knowing the limits in $(2)$ and using limit theorems. You have one addition ($1 + \cos x$) followed by a division ($/(1 + \cos x)$) and a multiplication. In each step you can start with an $\epsilon$ and find a corresponding $\delta$ (like in the proof of corresponding limit theorem) and thereby finally obtain a $\delta$ for a given $\epsilon$ which is sufficient to prove $(1)$.
Note: You can simplify the above process if you use following equation instead of $(3)$: $$\frac{1 - \cos x}{x^{2}} = \frac{1}{2}\cdot\frac{\sin (x/2)}{x/2}\cdot\frac{\sin (x/2)}{x/2}\tag{4}$$
But doing this does not help much except for being a cumbersome exercise. It will be best to study and understand the proofs of limit theorems and then apply them in evaluating limits.
We need to find a δ so that if $$|x| < \delta$$
$$|\frac{1 - cos(x)}{x ^ 2} - \frac{1}{2}| < \varepsilon$$
Note that
$$|\frac{1 - cos(x)}{x ^ 2}| = |\frac{1 - cos ^ 2(x)}{x ^ 2(1 + cos(x)) }| = |\frac{sin^2(x)}{x ^ 2(1 + cos(x))}| = |\frac{sin(x)}{x}|^2|\frac{1}{1 + cos(x)}|$$
But $$\big|\hspace{1mm}|\frac{sin(x)}{x}|^2|\frac{1}{1 + cos(x)}| - \frac{1}{2}\big|$$ is an expression of the form $xy - x_0y_0$ where $$x = |\frac{sin(x)}{x}|^2\\ y = |\frac{1}{1 + cos(x)}|\\ x_0 = 1\\ y_0 = \frac{1}{2} $$
Since $$|xy - x_0y_0| = |(x - x_0)(y - y_0) + x(y - y_0) + y(x - x_0)| \\ \hspace{2.5cm}\leqslant |(x - x_0)(y - y_0)| + |x(y - y_0)| + |y(x - x_0)|$$
We need to find a $\delta$ such that all the three terms in the last expression are less that $\frac{\varepsilon}{3}$
Since $lim_{x\to0} y_0 = y$ and $lim_{x\to0} x_0 = x$, minimize each term separately and pick the smallest $\delta$ of the three you get. The result follows.
P.S. - This is my first time typing an answer out in latex. If I've made any errors formatting this answer, I'm sorry.
