Technique to solve limits

Question

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

$$\lim\limits_{x\to{-1}} \frac{x^3 + 1}{x^2 + 4x + 3}$$

Answer 1

This is practice, a lot...and observing carefully what $\;x\;$ is tending to, in this case $\;x\to-1\;$ .

Since you have a rational function here, it is continuous at any point where the denominator doesn't vanish, so one "suspects" $\;x=-1\;$ is a root of the denominator, otherwise the limit is obtained simply by substitution (substitute $\;x=-1\;$ into the function) , by continuity of the function.

Once you checked $\;-1\;$ indeed is a root of the denominator, then either it is a root of the numerator or not. You check, and you discover it actually is. Thus, as other answer mentioned, both polynomials above and below are divided by $\;x+1\;$ and etc.:

$$\frac{x^3+1}{x^2+4x+3}=\frac{\color{red}{(x+1)}(x^2-x+1)}{\color{red}{(x+1)}(x+3)}=\frac{x^2-x+1}{x+3}\overbrace{\xrightarrow[x\to-1]{}}^{\text{just substitute, by cont.!}}\frac{3}{2}$$

Answer 2

You know that by noting that $f(x) = x^3+1, g(x) = x^2+4x+3$ both are polynomials and have the property $f(-1) = 0 = g(-1)$, hence they have a common factor $x+1$.

Answer 3

You know these things by experience. Because when we start solving questions, we start to get the hang of it. Now when you do this example, you divide it by the expression, because you would have done factorising exercises before and you remember them.

Answer 4

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

$$\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{p(a)}{q(a)}\lim_{x\to a}(x-a)^{n-d}.$$

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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 $n=d$.