15. Find the following limit V2x+5–1 lim x→-2 x+2
Trigonometry (11th Edition)
11th Edition
ISBN:9780134217437
Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Chapter1: Trigonometric Functions
Section: Chapter Questions
Problem 1RE:
1. Give the measures of the complement and the supplement of an angle measuring 35°.
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![### Calculus Problem: Evaluating a Limit
**Problem:**
15. Find the following limit
\[ \lim_{{x \to -2}} \frac{\sqrt{2x + 5} - 1}{x + 2} \]
### Steps to Solve the Limit:
1. **Direct Substitution:**
Start by substituting \( x = -2 \) into the function to check if the limit can be directly evaluated:
\[ \frac{\sqrt{2(-2) + 5} - 1}{-2 + 2} = \frac{\sqrt{-4 + 5} - 1}{0} = \frac{\sqrt{1} - 1}{0} = \frac{1 - 1}{0} = \frac{0}{0} \]
Direct substitution results in an indeterminate form \(\frac{0}{0}\), which indicates that we need to use another method to evaluate the limit.
2. **Rationalizing the Numerator:**
To resolve this, rationalize the numerator by multiplying the numerator and denominator by the conjugate of the numerator \(\sqrt{2x + 5} + 1\):
\[ \frac{\sqrt{2x + 5} - 1}{x + 2} \cdot \frac{\sqrt{2x + 5} + 1}{\sqrt{2x + 5} + 1} \]
The goal is to simplify the expression:
\[ \frac{(\sqrt{2x + 5} - 1)(\sqrt{2x + 5} + 1)}{(x + 2)(\sqrt{2x + 5} + 1)} \]
Using the difference of squares \( (a - b)(a + b) = a^2 - b^2 \):
\[ = \frac{(2x + 5) - 1}{(x + 2)(\sqrt{2x + 5} + 1)} \]
Simplify the numerator:
\[ = \frac{2x + 4}{(x + 2)(\sqrt{2x + 5} + 1)} \]
Factor the numerator:
\[ = \frac{2(x + 2)}{(x + 2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F63120590-018c-4efa-893a-21ed83fdf24a%2F8a2631a6-b7ae-49ef-92cf-57b32a7ef592%2Fci8yie_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Calculus Problem: Evaluating a Limit
**Problem:**
15. Find the following limit
\[ \lim_{{x \to -2}} \frac{\sqrt{2x + 5} - 1}{x + 2} \]
### Steps to Solve the Limit:
1. **Direct Substitution:**
Start by substituting \( x = -2 \) into the function to check if the limit can be directly evaluated:
\[ \frac{\sqrt{2(-2) + 5} - 1}{-2 + 2} = \frac{\sqrt{-4 + 5} - 1}{0} = \frac{\sqrt{1} - 1}{0} = \frac{1 - 1}{0} = \frac{0}{0} \]
Direct substitution results in an indeterminate form \(\frac{0}{0}\), which indicates that we need to use another method to evaluate the limit.
2. **Rationalizing the Numerator:**
To resolve this, rationalize the numerator by multiplying the numerator and denominator by the conjugate of the numerator \(\sqrt{2x + 5} + 1\):
\[ \frac{\sqrt{2x + 5} - 1}{x + 2} \cdot \frac{\sqrt{2x + 5} + 1}{\sqrt{2x + 5} + 1} \]
The goal is to simplify the expression:
\[ \frac{(\sqrt{2x + 5} - 1)(\sqrt{2x + 5} + 1)}{(x + 2)(\sqrt{2x + 5} + 1)} \]
Using the difference of squares \( (a - b)(a + b) = a^2 - b^2 \):
\[ = \frac{(2x + 5) - 1}{(x + 2)(\sqrt{2x + 5} + 1)} \]
Simplify the numerator:
\[ = \frac{2x + 4}{(x + 2)(\sqrt{2x + 5} + 1)} \]
Factor the numerator:
\[ = \frac{2(x + 2)}{(x + 2
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