How can I compute $\lim_{x\rightarrow-1}\frac{x^2+3x+2}{x^2+4x+3}$, $\lim_{x\rightarrow 0^{+}}\frac{x^2-3x+2}{x^3-2x^2}$

Question

I've checked out both of these functions at the points, where I have to compute the limits of the functions. I assume that $\lim_{x\rightarrow-1}\frac{x^2+3x+2}{x^2+4x+3}$ does not exist, but how can I compute it? How can I compute the one-side limit of the second function?

Answer 1

note that $$\frac{x^2+3x+2}{x^2+4x+3}=\frac{(x+2)(x+1)}{(x+3)(x+1)}$$ and write $$\frac{x^2-3x+2}{x^2(x-2)}$$ Can you finish? for the second example we get $$-\infty$$

Answer 2

For what concerns the second one, since the first one has already been solved by Dr. Sonnhard G., you simply notice that

$$\frac{x^2-3x+2}{x^3-2x^2} = \frac{(x-2)(x-1)}{x^2(x-2)}$$

Now you can simplify the identical terms to get

$$\frac{x-1}{x^2}$$

Hence the second one goes to $-\infty$