Calculate $\lim\limits_{x\to -1}\frac{x^2+3x+2}{x^2+2x+1}$ I've just have a mathematics exam and a question was this: 
Calculate the limits of $\dfrac{x^2+3x+2}{x^2+2x+1}$ when $x\text{ aproaches }-1$.
I started by dividing it using the polynomial long division. But I always get $\frac{0}{0}$.
How is this limit evaluated?

Hint for an alternate method: ${x^2+3x+2\over x^2+2x+1}={(x+2)(x+1)\over(x+1)(x+1)} $. (After cancelling, look at the limits from the left and right.)

Long division will get you the result after a bit more work as well. Doing the division reveals
$$
{x^2+3x+2\over x^2+2x+1} =1+{x+1\over x^2+2x+1}.
$$
Now, using $x^2+2x+1=(x+1)^2$, write this as
$$
1+{1\over x+1}.
$$
And then consider $\lim\limits_{x\rightarrow -1}\Bigl(1+{1\over x+1}\Bigr)$. (I think the first method above is a bit less work.)
Can you handle this now?
