Find expert answers and community insights on IDNLearn.com. Discover in-depth answers to your questions from our community of experienced professionals.
Sagot :
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]
### 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]
We appreciate your contributions to this forum. Don't forget to check back for the latest answers. Keep asking, answering, and sharing useful information. Find the answers you need at IDNLearn.com. Thanks for stopping by, and come back soon for more valuable insights.
