Problem 74 - Precalculus with Limits

Question

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

Step-by-Step Solution

Answer

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 1: 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: \(\frac{\sqrt{(x+h)-2} - \sqrt{x-2}}{h}\)

Step 2: 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: \(\frac{1}{\sqrt{(x+h)-2} + \sqrt{x-2}}\).

Step 3: Substitute \(h\) with 0

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

Key Concepts

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.