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?
If you divided correctly, you got $$1+\frac{x+1}{x^2+2x+1}\;.\tag{1}$$ You should recognize that the denominator is simply $(x+1)^2$, so $(1)$ is $$1+\frac{x+1}{(x+1)^2}=1+\frac1{x+1}\;.$$ Thus, your limit is $$\lim_{x\to-1}\left(1+\frac1{x+1}\right)=1+\lim_{x\to-1}\frac1{x+1}\;,$$ and you shouldn't have too much trouble seeing what's going on with that last limit.
Had you recognized that the original numerator is $(x+1)(x+2)$, you could have divided out the common factor of $x+1$ and reached this stage somewhat more quickly.
Another approach is to exploit the fact that it's easier to compute limits of rational functions at $0$. Thus changing variables $\rm\:z = x\!+\!1\:$ shifts $\rm\:x = -1\:$ to $\rm\:z = 0,\:$ so with $\rm\:x = z\!-\!1\:$ we get
$$\rm\ \lim_{z\to 0} \frac{(z\!-\!1)^2+3(z\!-\!1)+2}{(z\!-\!1)^2+2(z\!-\!1)+1}\ =\ \lim_{z\to 0}\frac{z^2 + z}{z^2}\ =\ \lim_{z\to 0}\:\left(1 + \frac{1}z\right)$$
You can do also this way,
Using l'Hospital rule,
$\lim_{x \to -1} \frac{x^2+3x+2}{x^2+2x+1} \ \ \ \ \ \ (\frac{0}{0}form)$
Applying L'hospital rule,
$\lim_{x \to -1} \frac{2x+3}{2x+2}$
$\lim_{x \to -1} \frac{2x+2+1}{2x+2}$
$\lim_{x \to -1} 1+\frac{1}{2x+2}$
can you do this now?
