To solve the limit problem [tex]\(\lim _{x \rightarrow -3} \frac{x^3 + 4x^2 + 4x + 3}{x^2 + 2x - 3}\)[/tex], let's proceed through the following steps:
1. Identify the expressions in the numerator and the denominator:
- The numerator is [tex]\( f(x) = x^3 + 4x^2 + 4x + 3 \)[/tex].
- The denominator is [tex]\( g(x) = x^2 + 2x - 3 \)[/tex].
2. Factorize the denominator:
- The denominator can be factorized as:
[tex]\[
x^2 + 2x - 3 = (x + 3)(x - 1)
\][/tex]
3. Factorize the numerator (if possible):
- To factorize the numerator [tex]\( x^3 + 4x^2 + 4x + 3 \)[/tex], we use polynomial factorization techniques such as synthetic division or the Rational Root Theorem. Upon factorizing, we find it can be written as:
[tex]\[
x^3 + 4x^2 + 4x + 3 = (x + 3)(x^2 + x + 1)
\][/tex]
4. Simplify the fraction:
- Substituting the factorizations into the original expression:
[tex]\[
\frac{x^3 + 4x^2 + 4x + 3}{x^2 + 2x - 3} = \frac{(x + 3)(x^2 + x + 1)}{(x + 3)(x - 1)}
\][/tex]
- We can cancel out the common factor [tex]\( (x + 3) \)[/tex], assuming [tex]\( x \neq -3 \)[/tex]:
[tex]\[
\frac{(x^2 + x + 1)}{(x - 1)}
\][/tex]
5. Evaluate the limit as [tex]\( x \)[/tex] approaches [tex]\(-3\)[/tex]:
- Now we need to find the limit of [tex]\( \frac{x^2 + x + 1}{x - 1} \)[/tex] as [tex]\( x \)[/tex] approaches [tex]\(-3\)[/tex]:
[tex]\[
\lim _{x \rightarrow -3} \frac{x^2 + x + 1}{x - 1}
\][/tex]
- Substitute [tex]\( x = -3 \)[/tex] in the simplified expression:
[tex]\[
\frac{(-3)^2 + (-3) + 1}{-3 - 1} = \frac{9 - 3 + 1}{-4} = \frac{7}{-4} = -\frac{7}{4}
\][/tex]
Therefore, the limit is:
[tex]\[
\lim _{x \rightarrow -3} \frac{x^3 + 4x^2 + 4x + 3}{x^2 + 2x - 3} = -\frac{7}{4}
\][/tex]
