Problem 20 - Calculus

Question

Estimate the limit by substituting smaller and smaller values of \(h .\) For trigonometric functions, use radians. Give answers to one decimal place. $$\lim _{h \rightarrow 0} \frac{\sqrt{1+h}-1}{h}$$

Step-by-Step Solution

Answer

The limit is approximately 0.5.

Step 1: Understanding the Problem

We need to find the limit as \(h\) approaches 0 of \(\frac{\sqrt{1+h}-1}{h}\). This requires us to evaluate the expression for smaller and smaller values of \(h\).

Step 2: Substitute Small Values of h

Start by substituting small values for \(h\). For instance, try \(h = 0.1, 0.01, 0.001\), and calculate \(\frac{\sqrt{1+h}-1}{h}\).

Step 3: Calculate for h = 0.1

Substitute \(h = 0.1\) into the expression:\[\frac{\sqrt{1+0.1}-1}{0.1} = \frac{\sqrt{1.1}-1}{0.1}\]Using a calculator, \(\sqrt{1.1} \approx 1.0488\). Thus, the expression becomes:\[\frac{1.0488-1}{0.1} \approx 0.488\]

Step 4: Calculate for h = 0.01

Substitute \(h = 0.01\) into the expression:\[\frac{\sqrt{1+0.01}-1}{0.01} = \frac{\sqrt{1.01}-1}{0.01}\]Using a calculator, \(\sqrt{1.01} \approx 1.00499\). Thus, the expression becomes:\[\frac{1.00499-1}{0.01} \approx 0.499\]

Step 5: Calculate for h = 0.001

Substitute \(h = 0.001\) into the expression:\[\frac{\sqrt{1+0.001}-1}{0.001} = \frac{\sqrt{1.001}-1}{0.001}\]Using a calculator, \(\sqrt{1.001} \approx 1.000499875\). Thus, the expression becomes:\[\frac{1.000499875-1}{0.001} \approx 0.4999\]

Step 6: Analyze the Pattern

From the calculations, as \(h\) gets smaller, the values \(0.488, 0.499, 0.4999\) come closer to \(0.5\). This suggests our estimated limit is near \(0.5\).

Step 7: Conclusion

Therefore, through the substitution of progressively smaller \(h\), the limit of \(\frac{\sqrt{1+h}-1}{h}\) as \(h\) approaches 0 is approximately \(0.5\).

Key Concepts

Trigonometric Functions

Trigonometric functions are fundamental in mathematics, often used to model periodic phenomena. However, in limit problems, they can also appear in more unexpected contexts. When dealing with trigonometric limits, remember to use radians instead of degrees since radians make the calculations more straightforward and intuitive. Unlike degree measures, which are based on arbitrary divisions, radians relate directly to the lengths of arcs, allowing for seamless integrations and derivations of trigonometric functions. It's also important to recognize certain linearizations and approximations, like how \(\sin(x) \approx x\) when \(x\) is near zero. While this exercise doesn't explicitly involve trigonometric functions, the methodology used here is similar: observing behavior as a variable gets infinitesimally small.

Approaching Zero

When we say a variable is "approaching zero," it means we are interested in its behavior as it becomes very close to, but never actually reaching, zero. This is fundamental in calculus as limits explore how expressions behave near certain points without actually being at those points.
Calculating limits like \(\lim_{h \to 0} \frac{\sqrt{1+h} - 1}{h}\) involves considering what happens as \(h\) goes toward zero. Our expression simplifies when \(h\) is very small, exhibiting predictable behaviors that might not be obvious at first glance.
The derivation is: \(\lim_{h \to 0} \frac{\sqrt{1+h} - 1}{h} = \frac{1}{2}\) by recognizing the derivative of the square root function. However, here the focus is getting the approximate understanding through numerical substitution.

Numerical Substitution

Numerical substitution is a practical technique to estimate limits, especially when an exact algebraic manipulation seems complex. By substituting progressively smaller values of \(h\), we observe how the function's value tends to stabilize towards a specific number. This method is particularly useful for educational purposes, as it visually demonstrates the concept of limits without requiring complex calculations.

• Calculate for \(h = 0.1\): \(\frac{\sqrt{1.1}-1}{0.1} \approx 0.488\)
• For \(h = 0.01\): \(\frac{\sqrt{1.01}-1}{0.01} \approx 0.499\)
• And for \(h = 0.001\): \(\frac{\sqrt{1.001}-1}{0.001} \approx 0.4999\)

As \(h\) decreases, the computed values increase toward 0.5, thus estimating that the limit is close to 0.5.
While numerical substitution is insightful, it's crucial to remember it gives an approximation rather than an exact value. But it's a powerful tool to foster intuitive understanding of limits and continuity.