Problem 22
Question
Find the limits in Exercises \(11-22\) $$\lim _{h \rightarrow 0} \frac{\sqrt{5 h+4}-2}{h}$$
Step-by-Step Solution
Verified Answer
The limit is \( \frac{5}{4} \).
1Step 1: Recognize the Form
The given limit is \( \lim _{h \rightarrow 0} \frac{\sqrt{5h+4} - 2}{h} \). This is an indeterminate form 0/0, which suggests using algebraic techniques to simplify it.
2Step 2: Multiply by the Conjugate
To eliminate the square root, multiply and divide by the conjugate to simplify:\[\lim _{h \rightarrow 0} \frac{\sqrt{5h+4} - 2}{h} \cdot \frac{\sqrt{5h+4} + 2}{\sqrt{5h+4} + 2} = \lim _{h \rightarrow 0} \frac{(\sqrt{5h+4})^2 - 2^2}{h(\sqrt{5h+4} + 2)}\]
3Step 3: Simplify the Numerator
Simplify the numerator using the difference of squares formula:\[(\sqrt{5h+4})^2 - 2^2 = 5h + 4 - 4 = 5h.\] This reduces the expression to:\[\lim _{h \rightarrow 0} \frac{5h}{h(\sqrt{5h+4} + 2)}.\]
4Step 4: Cancel the Common Factor
Cancel \(h\) from the numerator and denominator:\[\lim _{h \rightarrow 0} \frac{5}{\sqrt{5h+4} + 2}.\]
5Step 5: Evaluate the Limit
Substitute \(h = 0\) in the simplified expression:\[\lim _{h \rightarrow 0} \frac{5}{\sqrt{5(0)+4} + 2} = \lim _{h \rightarrow 0} \frac{5}{2 + 2} = \frac{5}{4}.\]
6Step 6: Final Answer
The limit is \( \frac{5}{4} \).
Key Concepts
Indeterminate FormsAlgebraic TechniquesConjugate Method
Indeterminate Forms
An indeterminate form often appears as 0/0 or ∞/∞ when evaluating limits. These forms signal that straightforward substitution into a limit won't work and require further analysis. In our exercise, the limit expression \( \lim _{h \rightarrow 0} \frac{\sqrt{5h+4} - 2}{h} \) results in 0/0 if you directly substitute \( h = 0 \).
This is a common situation in calculus where expressions approach zero in both the numerator and the denominator. But, this doesn't mean the limit doesn't exist. Instead, it implies that more manipulation is needed to resolve the form and find a meaningful limit.
Understanding indeterminate forms is crucial as they open up opportunities to apply various algebraic techniques to simplify the problem and arrive at a definite answer. It's like the universe providing a clue that there’s more beneath the surface.
This is a common situation in calculus where expressions approach zero in both the numerator and the denominator. But, this doesn't mean the limit doesn't exist. Instead, it implies that more manipulation is needed to resolve the form and find a meaningful limit.
Understanding indeterminate forms is crucial as they open up opportunities to apply various algebraic techniques to simplify the problem and arrive at a definite answer. It's like the universe providing a clue that there’s more beneath the surface.
Algebraic Techniques
Algebraic techniques come to the rescue when dealing with indeterminate forms. These methods help transform difficult expressions into more manageable forms. In our example, we use the concept of conjugates to tackle the indeterminate form 0/0.
One common technique is the "Difference of Squares". It's employed when we have expressions that include square roots and their conjugates. Multiplying the numerator and the denominator by the conjugate, \( \sqrt{5h+4} + 2 \), we create a "difference of squares" scenario:
One common technique is the "Difference of Squares". It's employed when we have expressions that include square roots and their conjugates. Multiplying the numerator and the denominator by the conjugate, \( \sqrt{5h+4} + 2 \), we create a "difference of squares" scenario:
- This simplifies our original expression significantly.
- It helps eliminate the square root.
Conjugate Method
The conjugate method involves multiplying an expression by its conjugate to eliminate square roots or other complicated elements. It’s particularly powerful in calculus when dealing with complex radicals.
In the expression \( \lim _{h \rightarrow 0} \frac{\sqrt{5h+4} - 2}{h} \), we multiply the numerator and the denominator by \( \sqrt{5h+4} + 2 \), the conjugate of \( \sqrt{5h+4} - 2 \). The product forms a difference of squares:\[\frac{(\sqrt{5h+4})^2 - 2^2}{h(\sqrt{5h+4} + 2)}\]
This subtraction simplifies the square roots, leaving us only algebraic expressions to handle. After that, the limit becomes easy to evaluate.
The key takeaway from the conjugate method is its ability to simplify radical expressions systematically. By transforming the indeterminate form into a solvable limit, the conjugate method becomes a trusted friend in the toolkit of calculus techniques. Understanding this method can significantly ease the process of finding limits.
In the expression \( \lim _{h \rightarrow 0} \frac{\sqrt{5h+4} - 2}{h} \), we multiply the numerator and the denominator by \( \sqrt{5h+4} + 2 \), the conjugate of \( \sqrt{5h+4} - 2 \). The product forms a difference of squares:\[\frac{(\sqrt{5h+4})^2 - 2^2}{h(\sqrt{5h+4} + 2)}\]
This subtraction simplifies the square roots, leaving us only algebraic expressions to handle. After that, the limit becomes easy to evaluate.
The key takeaway from the conjugate method is its ability to simplify radical expressions systematically. By transforming the indeterminate form into a solvable limit, the conjugate method becomes a trusted friend in the toolkit of calculus techniques. Understanding this method can significantly ease the process of finding limits.
Other exercises in this chapter
Problem 22
Find the limits in Exercises \(21-42\) $$\lim _{t \rightarrow 0} \frac{\sin k t}{t} \quad(k \text { constant })$$
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Each of Exercises \(15-30\) gives a function \(f(x)\) and numbers \(L, c,\) and \(\epsilon>0 .\) In each case, find an open interval about \(c\) on which the in
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At what points are the functions in Exercises 13-30 continuous? $$y=\frac{x \tan x}{x^{2}+1}$$
View solution Problem 23
The process by which we determine limits of rational functions applies equally well to ratios containing noninteger or negative powers of \(x :\) Divide numerat
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