Know Your Limits

We will simplify the expression inside the root. $1-cos2(x-2) = 2sin^2(x-2)$ , therefore $\sqrt{2sin^2(x-2)} = \sqrt{2}|sin(x-2)|$ . The question wants us to find $\lim_{x \rightarrow 2} \frac{\sqrt{2}|sin(x-2)|}{x-2}$

In general $\lim_{x \rightarrow 0} \frac{sin(x)}{x} = 1$ for both left-handed and right-handed limit, but when it involved the absolute value, there will be the deviation as regardless of the value of $x$ , $|sin(x)|$ will always give positive value while $x$ itself may not, thus: $\lim_{x \rightarrow 0^-} \frac{|sin(x)|}{x} = -1; \lim_{x \rightarrow 0^+} \frac{|sin(x)|}{x} = 1$ In similar vein $\lim_{x \rightarrow 2^-} \frac{\sqrt{2}|sin(x-2)|}{x-2} = -\sqrt{2}; \lim_{x \rightarrow 2^+} \frac{\sqrt{2}|sin(x-2)|}{x-2} = \sqrt{2}$ therefore $\lim_{x \rightarrow 2} \frac{\sqrt{1-cos2(x-2)}}{x-2}$ is not determined as both sides of the limit yields different result

I utilized L'Hopital's Rule and that yields 0/0.