Certainly! Let's find the limit of the function [tex]\( \frac{2x^2 + 9x - 5}{x + 5} \)[/tex] as [tex]\( x \)[/tex] approaches -5.
### Step-by-Step Solution:
1. Identify the problem:
We need to evaluate the limit:
[tex]\[
\lim_{{x \to -5}} \frac{2x^2 + 9x - 5}{x + 5}
\][/tex]
2. Direct substitution:
First, try substituting [tex]\( x = -5 \)[/tex] directly into the function. If direct substitution does not result in an indeterminate form (like [tex]\( \frac{0}{0} \)[/tex]), then the limit is simply the value of the function at that point.
[tex]\[
\frac{2(-5)^2 + 9(-5) - 5}{-5 + 5}
\][/tex]
[tex]\[
= \frac{2(25) - 45 - 5}{0}
\][/tex]
[tex]\[
= \frac{50 - 45 - 5}{0}
\][/tex]
[tex]\[
= \frac{0}{0}
\][/tex]
Since substituting [tex]\( x = -5 \)[/tex] results in the indeterminate form [tex]\( \frac{0}{0} \)[/tex], we cannot determine the limit directly.
3. Factorize the numerator:
To resolve the indeterminate form, we factorize the numerator and cancel out the common factor.
[tex]\[
2x^2 + 9x - 5
\][/tex]
Let's factorize this quadratic expression.
Solving the quadratic equation:
[tex]\( 2x^2 + 9x - 5 = 0 \)[/tex]
Using the factorization method, we find that:
[tex]\[
2x^2 + 9x - 5 = (2x - 1)(x + 5)
\][/tex]
4. Rewrite the limit:
Now, we rewrite the limit expression using the factorized form of the numerator:
[tex]\[
\lim_{{x \to -5}} \frac{(2x - 1)(x + 5)}{x + 5}
\][/tex]
Cancel out the common factor [tex]\((x + 5)\)[/tex] in the numerator and the denominator:
[tex]\[
\lim_{{x \to -5}} \frac{(2x - 1) \cancel{(x + 5)}}{\cancel{(x + 5)}}
\][/tex]
[tex]\[
= \lim_{{x \to -5}} (2x - 1)
\][/tex]
5. Direct substitution:
Now, substitute [tex]\( x = -5 \)[/tex] in the simplified expression [tex]\( 2x - 1 \)[/tex]:
[tex]\[
2(-5) - 1
\][/tex]
[tex]\[
= -10 - 1
\][/tex]
[tex]\[
= -11
\][/tex]
### Conclusion:
Therefore, the limit of [tex]\( \frac{2x^2 + 9x - 5}{x + 5} \)[/tex] as [tex]\( x \)[/tex] approaches -5 is:
[tex]\[
\boxed{-11}
\][/tex]
