How to graph $f(x)=(1-\cos(2x-2))/(x-1)^2$ analytically?

Question

I am asked to graph the function

$$f(x)=\frac{1-\cos(2x-2)}{(x-1)^2}$$

I only have an issue when it comes to determining the hole at $x=1$ analytically. Graphing the function on a graphing utility reveals a hole at $x=1$ rather than a vertical asymptote. I know that by simply plugging in the value $1$ for $x$, one will see that both the numerator and denominator equate to zero. This shows that there is at least one factor of $(x-1)$ that can be extracted from the numerator.

I would like to know how to simplify this function analytically in such a way that two factors of $(x-1)$ can be obtained from the numerator. I believe one must make use of an inverse function down the line in order to extract the obvious $(x-1)$ hiding in the $\cos(2x-2)=\cos(2(x-1))$ term. That is as far as I got in the problem.

Thank you for your help.

Answer 1

Hint:   look at $\,f(x+1)=\dfrac{1- \cos 2x}{x^2} = \dfrac{2 \sin^2 x}{x^2}\,$, then shift the origin back.