Problem 62 - Thomas Calculus

Question

Because of their connection with secant lines, tangents, and instantaneous rates, limits of the form $$\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$$ occur frequently in calculus. In Exercises \(57-62,\) evaluate this limit for the given value of \(x\) and function \(f\) . $$f(x)=\sqrt{3 x+1}, \quad x=0$$

Step-by-Step Solution

Answer

The limit is \( \frac{3}{2} \).

Step 1: Set up the Limit Expression

The given function is \( f(x) = \sqrt{3x + 1} \) and we need to evaluate the limit \( \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h} \) for \( x = 0 \). First, substitute \( f(x) \) and \( f(x+h) \) into the limit to get \( \lim_{h \rightarrow 0} \frac{\sqrt{3(0+h) + 1} - \sqrt{3(0) + 1}}{h} \), which simplifies to \( \lim_{h \rightarrow 0} \frac{\sqrt{3h + 1} - 1}{h} \).

Step 2: Use the Conjugate to Simplify

To simplify the expression, multiply the numerator and the denominator by the conjugate of the numerator, which is \( \sqrt{3h + 1} + 1 \). This gives: \[ \lim_{h \rightarrow 0} \frac{(\sqrt{3h + 1} - 1)(\sqrt{3h + 1} + 1)}{h(\sqrt{3h + 1} + 1)} \].

Step 3: Simplify the Numerator

Apply the difference of squares in the numerator: \( (\sqrt{3h + 1})^2 - 1^2 = (3h + 1) - 1 = 3h \). The limit is now \( \lim_{h \rightarrow 0} \frac{3h}{h(\sqrt{3h + 1} + 1)} \).

Step 4: Cancel the Common Term

Cancel the \( h \) terms from the numerator and the denominator: \( \lim_{h \rightarrow 0} \frac{3}{\sqrt{3h + 1} + 1} \).

Step 5: Evaluate the Limit

Directly substitute \( h = 0 \) into the simplified expression: \( \frac{3}{\sqrt{3(0) + 1} + 1} = \frac{3}{2} \). Thus, the limit is \( \frac{3}{2} \).

Key Concepts

Limits

Limits are fundamental to calculus and are crucial in understanding how functions behave as they approach specific points. The limit we evaluate here reflects the derivative concept, which geometrically represents the slope of the tangent line to the curve at a point.
To understand this, think of limits as a way to explore what happens to a function's value as the input gets infinitely close to a certain number.

• Notation: The notation \( \lim_{h \rightarrow 0} \) means you are interested in what happens to your expression as \( h \) becomes very small, close to zero.
• Purpose: In the context of derivatives, limits help us find the precise rate of change of a function at a single point.

Recognizing the importance of limits allows us to deal with indeterminate forms like \( \frac{0}{0} \), often encountered in calculus. For this, simplifying expressions is key, which brings us to our next concept.

Difference Quotient

The difference quotient is the backbone of derivative calculations. It's the expression \( \frac{f(x+h) - f(x)}{h} \) that we simplify using limits to find a derivative.
This represents the slope of the secant line—a line through two points on the graph of the function—and becomes the tangent slope when \( h \) approaches zero.

• Expression: It's a ratio of differences. The numerator \( f(x+h) - f(x) \) highlights the change in the function values, while \( h \) in the denominator represents the change in \( x \).
• Meaning: As \( h \rightarrow 0 \), we capture how sharply \( f(x) \) changes at a point \( x \), leading directly to the derivative \( f'(x) \).

Through this quotient, typically, you measure how a curve tilts at a point, shifting from average changes to instantaneously precise changes. This shift gives key insight into a function's behavior.

Conjugate Multiplication

Conjugate multiplication is a powerful algebraic technique used to simplify complex expressions, particularly those involving square roots.
In the exercise, the use of conjugates helps rationalize the difference quotient by exploiting the difference of squares formula.

• Technique: Multiply the numerator and denominator by the conjugate of the numerator, which converts a complex expression into a simpler, rational form.
• Benefit: This step transforms the quotient from a potentially indeterminate form \( \frac{0}{0} \) to one that can be evaluated using direct substitution.

The rationalizing process exposes algebraic simplifications to resolve the expression into a workable form. Through these elegant manipulations, conjugate multiplication not only clarifies expressions but also smoothens the path to evaluating the limit.