Technique to solve limits I was making list of limits exercises, I can't use L' hôpital to solve, I have to solve using only the properties of limits. The only techniques that I know are: I. trying to replace x by the number II. divide III. multiply under the terms of the conjugate The following limit is solved by dividing x + 1each term, the explanation is because both polynomials are divisible by x + 1 but how do I know that? I tried to divide the terms for x³ and x² and the answer has always zero, but the result is 3/2
$$\lim\limits_{x\to{-1}} \frac{x^3 + 1}{x^2 + 4x + 3}$$

You first identify the type of indeterminacy by evaluating the function. Here you notice that the numerator and denominator, which are polynomials, tend to zero, so that you have a case $\frac00$.
A first technique to resolve such indeterminacies is by factoring out the roots and counting their multiplicities.
If you can factor as $$\frac{f(x)}{g(x)}=\frac{p(x)(x-a)^n}{q(x)(x-a)^d}$$ with $p(a),q(a)\ne0$, you may simplify to get
$$\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{p(a)}{q(a)}\lim_{x\to a}(x-a)^{n-d}.$$

In the given case, factorization is easy as you already know the common root $x=-1$, and
$$\lim_{x\to-1}\frac{x^3+1}{x^2+4x+3}=\lim_{x\to-1}\frac{(x^2-x+1)(x+1)}{(x+3)(x+1)}=\frac{(-1)^2-(-1)+1}{(-1)+3}
\cdot1,$$as $n=d$.
