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