Let's find the limit
[tex]\[
\lim_{{x \to 4^+}} \frac{2x + 1}{4 - x}
\][/tex]
Step-by-step:
1. Expression Setup:
We have the function [tex]\(\frac{2x + 1}{4 - x}\)[/tex] and we want to find its limit as [tex]\(x\)[/tex] approaches 4 from the right ([tex]\(x \to 4^+\)[/tex]).
2. Analyzing the Behavior as [tex]\(x \to 4^+\)[/tex]:
- Numerator Analysis:
The numerator [tex]\(2x + 1\)[/tex] when [tex]\(x\)[/tex] is slightly greater than 4 will be:
[tex]\[
2(4 + \epsilon) + 1 = 2 \cdot 4 + 2 \epsilon + 1 = 9 + 2 \epsilon
\][/tex]
which approaches 9 as [tex]\( \epsilon \)[/tex] (a small positive number) approaches 0.
- Denominator Analysis:
The denominator [tex]\(4 - x\)[/tex] when [tex]\(x\)[/tex] is slightly greater than 4 will be:
[tex]\[
4 - (4 + \epsilon) = -\epsilon
\][/tex]
which approaches 0 from the negative side as [tex]\( \epsilon \)[/tex] approaches 0.
3. Form of the Fraction:
Since the numerator approaches 9 and the denominator approaches 0 but is negative, the form of the fraction is:
[tex]\[
\frac{9 + 2 \epsilon}{-\epsilon}
\][/tex]
4. Behavior of the Fraction:
As [tex]\(\epsilon\)[/tex] approaches 0 from the positive side:
[tex]\[
\frac{9 + 2 \epsilon}{-\epsilon} \approx \frac{9}{-\epsilon}
\][/tex]
This fraction will become highly negatively infinite because [tex]\(\epsilon\)[/tex] is a very small positive number:
[tex]\[
\frac{9}{-\epsilon} \to -\infty \text{ as } \epsilon \to 0^+
\][/tex]
5. Conclusion:
Therefore, as [tex]\(x\)[/tex] approaches 4 from the right, [tex]\(\frac{2x + 1}{4 - x}\)[/tex] approaches [tex]\(-\infty\)[/tex].
So the limit is:
[tex]\[
\boxed{-\infty}
\][/tex]
