$\displaystyle x\neq {-3,1}$

a) $\displaystyle \lim_{x \to -3+}g(x) = \frac{5x^2}{x^2+2x-3}$

b) $\displaystyle \lim_{x \to 1-}g(x) = \frac{5x^2}{x^2+2x-3}$

I can solve these by subbing in values just a bit greater than 3 and using the same technique for b), but how do I solve these algebraically. There were two questions before these where $\displaystyle x~\rightarrow$ $\displaystyle \infty$ (same function) and I just divided the top and bottom by $\displaystyle x^2$ to get easily solvable limits. I tried this for these two and I get a denominator of 0 - so it doesn't work.