Problem 57
Question
In Exercises 49-68, find the limit by direct substitution. $$ \lim_{x \to -2}\ \dfrac{5x+3}{2x-9}$$
Step-by-Step Solution
Verified Answer
After carrying out the above steps, we find that the limit of the function as \( x \) approaches -2 is -1.
1Step 1: Substitution
The value of \( x \) given is \( -2 \). So, substitute \( x = -2 \) into the given expression \( \dfrac{5x+3}{2x-9} \)
2Step 2: Compute the Result
After substituting \( -2 \) for \( x \) , we will compute the numerator and the denominator separately (i.e calculate \( 5(-2)+3 \) and \( 2(-2)-9 \) respectively ), then divide to find the answer.
Key Concepts
Direct SubstitutionNumerator and DenominatorComputing Limits
Direct Substitution
Direct substitution is one of the simplest ways to compute limits. The method involves directly replacing the variable in the expression with the number it is approaching. In our exercise, we are dealing with the limit \( \lim_{x \to -2}\ \frac{5x+3}{2x-9} \). Here, \( x \) approaches \(-2\).
To use direct substitution, we substitute \( x = -2 \) directly into the expression. This method relies on the function being nice enough for this substitution nature, meaning it doesn't result in a division by zero or other undefined situations.
When successful, direct substitution efficiently provides the limit without further manipulation. It works particularly well when the function in question is continuous at the point of interest.
To use direct substitution, we substitute \( x = -2 \) directly into the expression. This method relies on the function being nice enough for this substitution nature, meaning it doesn't result in a division by zero or other undefined situations.
When successful, direct substitution efficiently provides the limit without further manipulation. It works particularly well when the function in question is continuous at the point of interest.
Numerator and Denominator
Understanding how the numerator and denominator separately affect the result is crucial when working with fractional expressions. In our expression \( \frac{5x+3}{2x-9} \), the numerator is \( 5x + 3 \) and the denominator is \( 2x - 9 \). Each portion has its effect on the outcome, and their interaction is essential.
When computing the limit, substitute \( x = -2 \) into both the numerator and denominator:
When computing the limit, substitute \( x = -2 \) into both the numerator and denominator:
- Numerator: \( 5(-2) + 3 = -10 + 3 = -7 \)
- Denominator: \( 2(-2) - 9 = -4 - 9 = -13 \)
Computing Limits
Once all substitutions and individual computations of the numerator and denominator are done, it's time to find the overall limit value. Calculating the final limit involves simply dividing the result of the numerator by the result of the denominator.
Through this step-by-step breakdown of direct substitution and individual calculations, you obtain your solution: \( \frac{7}{13} \). This process reaffirms how mathematical operations require precision and careful handling of expressions to achieve the correct result.
- Numerator value: \(-7\)
- Denominator value: \(-13\)
Through this step-by-step breakdown of direct substitution and individual calculations, you obtain your solution: \( \frac{7}{13} \). This process reaffirms how mathematical operations require precision and careful handling of expressions to achieve the correct result.
Other exercises in this chapter
Problem 57
In Exercises 55-58, find an equation of the line that is tangent to the graph of and parallel to the given line. \(\quad \quad Function \quad \quad \quad \quad
View solution Problem 57
In Exercises 55-62, graph the function. Determine the limit (if it exists) by evaluating the corresponding one-sided limits. $$\lim_{x \to 1} \dfrac{1}{x^2+1}$$
View solution Problem 58
In Exercises 55-58, find an equation of the line that is tangent to the graph of and parallel to the given line. \(\quad \quad Function \quad \quad \quad \quad
View solution Problem 58
In Exercises 55-62, graph the function. Determine the limit (if it exists) by evaluating the corresponding one-sided limits. $$\lim_{x \to 1} \dfrac{1}{x^2-1}$$
View solution