Prove $\lim_{x\to 0} \frac{ 1 -\cos x } {x^2} =\frac1 2$ Where did I go wrong?
(preferably a logical explanation instead of "this isn't how you use (insert name) rule)
\begin{align*}
&\lim_{x\to 0}\frac{ 1 -\cos x }{x²}\\
=&\lim_{x\to 0}\frac{ 1 -\cos x }{(x-0)\cdot(x-0)}\\
=&\lim_{x\to 0}\frac{\sin x }{x-0}\\
=&\lim_{x\to 0}\cos x\\
=&1.
\end{align*}
 A: The second equality doesn't hold. If you are using L'Hopital's rule, you should do\begin{align}\lim_{x\to0}\frac{1-\cos x}{x^2}&=\lim_{x\to0}\frac{\sin x}{2x}\\&=\frac12\lim_{x\to0}\frac{\sin x}x\\&=\frac12.\end{align}Another possibility is:\begin{align}\lim_{x\to0}\frac{1-\cos x}{x^2}&=\lim_{x\to0}\frac{2\sin^2\left(\frac x2\right)}{x^2}\\&=\frac12\lim_{x\to0}\frac{\sin^2\left(\frac x2\right)}{\left(\frac x2\right)^2}\\&=\frac12\left(\lim_{x\to0}\frac{\sin\left(\frac x2\right)}{\frac x2}\right)^2\\&=\frac12.\end{align}
A: Actually
$$\lim_{x \to 0} \frac{1-\cos x}{x\cdot x} = \lim_{x \to 0}\frac{\sin x}{2x}. $$
However, it is not necessary to use de l'Hôpital rule:$$\begin{align}\lim_{x \to 0} \frac{1-\cos x}{x^2}&=\lim_{x \to 0}\frac{1-\cos^2x}{x^2(1+\cos x)}\\ &= \lim_{x \to 0} \frac{\sin^2 x}{x^2}\frac{1}{1+\cos x} \\&= \frac{1}{2}.\end{align} $$
A: You have an indeterminate form.
Using L'Hopital's Rule, take the derivative of numerator and denominator, transforming your limit to:
\begin{align}\lim_{x\to0}\dfrac{1-\cos{x}}{x^2}=\lim_{x\to0}\dfrac{\sin{x}}{2x}\end{align}
This is another indeterminate form. Another iteration of L'Hopital's Rule implies:
\begin{align}\lim_{x\to0}\dfrac{\sin{x}}{2x} = \lim_{x\to0}\dfrac{\cos{x}}{2}\end{align}
Upon substitution of $x=0$,
\begin{align}\lim_{x\rightarrow0}\dfrac{\cos{x}}{2}=\dfrac{1}{2}\end{align}
