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Chapter 12: Problem 74

In Exercises 71-78, find \(\lim_{h \to 0}\ \dfrac{f(x+h)-f(x)}{h} \). \(f(x) = \sqrt{x-2}\)

Short Answer

Expert verified
The solution to the limit problem \(\lim_{h \to 0}\ \frac{f(x+h)-f(x)}{h}\) when \(f(x) = \sqrt{x-2}\) is \(\frac{1}{2\sqrt{x-2}}\).

Step by step solution

01

Apply the Difference Quotient Formula

Substitute the function \(f(x) = \sqrt{x-2}\) into the difference quotient formula: \(\frac{f(x+h)-f(x)}{h}\), which gives us: \(\frac{\sqrt{(x+h)-2} - \sqrt{x-2}}{h}\)
02

Rationalize the Numerator

To simplify this expression, we can rationalize the numerator by multiplying the numerator and denominator by the conjugate of the numerator: \(\frac{\sqrt{(x+h)-2} - \sqrt{x-2}}{h} * \frac{\sqrt{(x+h)-2} + \sqrt{x-2}}{\sqrt{(x+h)-2} + \sqrt{x-2}}\). This gives us: \(\frac{(x+h)-2 - (x-2)}{h[\sqrt{(x+h)-2} + \sqrt{x-2}]}\), which simplifies to: \(\frac{h}{h[\sqrt{(x+h)-2} + \sqrt{x-2}]}\). After cancelling out \(h\) in the numerator and denominator, we get: \(\frac{1}{\sqrt{(x+h)-2} + \sqrt{x-2}}\).
03

Substitute \(h\) with 0

Finally, substitute \(h\) with 0 in the simplified limit: \(\frac{1}{\sqrt{(x+0)-2} + \sqrt{x-2}} = \frac{1}{2\sqrt{x-2}}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Difference Quotient
The difference quotient is a central concept in calculus, especially when dealing with limits and derivatives. It is a formula that provides the average rate of change of a function over an interval. The formula is given by:
  • \( \frac{f(x+h)-f(x)}{h} \)
In this case, we are considering the function \(f(x) = \sqrt{x-2}\). The expression \(f(x+h)\) becomes \(\sqrt{(x+h)-2}\).
This transforms our difference quotient into:
  • \( \frac{\sqrt{(x+h)-2} - \sqrt{x-2}}{h} \)
As \(h\) approaches zero, the difference quotient provides an approximation of how the function behaves at a specific point. It acts as a stepping stone for finding the derivative, which is essentially the exact rate of change at a point.
Rationalizing the Numerator
Rationalizing the numerator is a technique used to simplify expressions that involve roots. By eliminating the roots in the numerator, we make it easier to evaluate limits.
The expression we start with is:
  • \( \sqrt{(x+h)-2} - \sqrt{x-2} \)
To rationalize, multiply both the numerator and the denominator by the conjugate of the numerator:
  • \( \sqrt{(x+h)-2} + \sqrt{x-2} \)
This clever trick uses the difference of squares formula, \((a-b)(a+b) = a^2 - b^2\), to remove the square roots:
  • \( \frac{h}{h[\sqrt{(x+h)-2} + \sqrt{x-2}]} \)
After rationalizing, we can cancel the common terms, simplifying the expression to a form that can be evaluated as \(h\) approaches zero.
Substitution Method
The substitution method is the final step in solving the limit problem. Once we have a simplified expression, we substitute \(h = 0\) to find the limit as \(h\) approaches zero.
  • The simplified expression is \( \frac{1}{\sqrt{(x+h)-2} + \sqrt{x-2}} \)
Substitute \(h = 0\) into the expression, which gives:
  • \( \frac{1}{\sqrt{x-2} + \sqrt{x-2}} \)
  • This further simplifies to \( \frac{1}{2\sqrt{x-2}} \)
By carefully analyzing the function's behavior as \(h\) becomes very small, substitution reveals the instantaneous rate of change.
This method solidifies the concept of a derivative, helping us see how small changes in \(x\) affect the function.

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Most popular questions from this chapter

In Exercises 9-28, find the limit (if it exists). If the limit does not exist, explain why. Use a graphing utility to verify your result graphically. \\[\lim_{t\to \infty} \dfrac{4t^2 - 2t + 1}{-3t^2 + 2t + 2} \\]

NUMERICAL AND GRAPHICAL ANALYSIS In Exercises 35-38, (a) complete the table and numerically estimate the limit as \(x\) approaches infinity, and (b) use a graphing utility to graph the function and estimate the limit graphically. $$ f(x) = x - \sqrt{x^2 + 2} $$

In Exercises 5-12, evaluate the sum using the summation formulas and properties. $$\displaystyle\sum_{k=1}^{20} (k^3 + 2)$$

In Exercises 9-28, find the limit (if it exists). If the limit does not exist, explain why. Use a graphing utility to verify your result graphically. \\[\lim_{x\to \infty} \dfrac{5x^3 + 1}{10x^3 - 3x^2 + 7} \\]

Use a graphing utility to graph the two functions given by $$ y_{1}=\frac{1}{\sqrt{x}} \text { and } y_{2}=\frac{1}{\sqrt[3]{x}} $$ in the same viewing window. Why does not appear to the left of the axis? How does this relate to the statement at the top of page 882 about the infinite limit $$ \lim _{x \rightarrow-\infty} \frac{1}{x^{r}} ? $$
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