I am having a hard time evaluating when a limit exists. I know that for the two-sided limit $\lim_{x\to a} f(x)$ to exist,both $\lim_{x\to a^+} f(x)$ and $\lim_{x\to a^-} f(x)$ must be equal, but it feels like I'm missing something.
For one,is it possible to evaluate this limit without choosing values on both sides of the limit and just "test it out"?
$\lim_{x\to -1} \dfrac{x^2+3x+1}{x^2+4x+3}$
You have $\lim_{x\to-1}x^2+3x+1=-1$ and $\lim_{x\to-1}x^2+4x+3=0$. Furthermore, since $x^2+4x+3=(x+1)(x+3)$, $x^2+4x+3>0$ if $x>-1$ and $x^2+4x+3<0$ if $x\in(-3,-1)$. Therefore,$$\lim_{x\to-1^+}\frac{x^2+3x+1}{x^2+4x+3}=-\infty\qquad\text{and}\qquad\lim_{x\to-1^-}\frac{x^2+3x+1}{x^2+4x+3}=\infty.$$It follows from this that your limit does not exist.