I am having trouble proving the continuity of:
$$f(x)=\frac{2x^5-98}{(x-9)(x-1)}$$
in the interval $[2,8]$.
I'm trying to use an epsilon-delta proof to get to the result but I keep getting stuck.
My proof so far looks like:
Suppose $|x-c|< \delta$ and let $\delta=1$.
\begin{align}|f(x)-f(c)| &= \left|\frac{2x^5-98}{(x-9)(x-1)}-\frac{2c^2-98}{(c-9)(c-1)}\right|\\\\ &=\left|\frac{(2x^5-98)(c-9)(c-1)-(2c^5-98)(x-9)(x-1)}{(x-9)(x-1)(c-9)(c-1)}\right|\\\\ &\leq \left|(2x^5-98)(c-9)(c-1)-(2c^5-98)(x-9)(x-1)\right|\\\\ &=\left|2x^2c^2(x^3-c^3)+20xc(x^4-y^4)+18(x^5-c^5)+98(x^2-c^2)+980(c-x)\right|\\ \end{align}
From here I think to use the triangle inequality and drag out a common factor of $|x-c|$ but even if I do that I still get stuck and don't know how to use my other assumption that $\delta=1$.
Am I on the right track or is there a more elegant way of showing continuity using epsilon-delta?