Problem 50 - Thomas Calculus

Question

Limits with trigonometric functions Find the limits in Exercises \(43-50 .\) $$\lim _{x \rightarrow 0} \sqrt{7+\sec ^{2} x}$$

Step-by-Step Solution

Answer

The limit is \(2\sqrt{2}\) as \(x\) approaches 0.

Step 1: Understanding the Problem

We need to find the limit \( \lim _{x \rightarrow 0} \sqrt{7+\sec ^{2} x} \). This involves evaluating the behavior of the function as \( x \) approaches 0.

Step 2: Simplify Inside the Square Root

Recall that \( \sec(x) = \frac{1}{\cos(x)} \), so \( \sec^2(x) = \frac{1}{\cos^2(x)} \). Substitute this expression in to rewrite the limit: \( \sqrt{7 + \sec^2 x} = \sqrt{7 + \frac{1}{\cos^2(x)}} \).

Step 3: Evaluate Known Limit Components

As \( x \rightarrow 0 \), \( \cos(x) \rightarrow 1 \), which implies \( \sec^2(x) = \frac{1}{\cos^2(x)} \rightarrow 1 \) as well.

Step 4: Plug in the Limit Approximations

Substitute the limiting behavior into the expression soinside the square root: \( \sqrt{7 + 1} = \sqrt{8} \).

Step 5: Finalize the Calculation

Hence, as \( x \rightarrow 0 \), the expression evaluates to \( \sqrt{8} \). To simplify further, \( \sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2} \).

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is an important and powerful tool for calculating limits that involve indeterminate forms. An indeterminate form arises when substituting a value into a limit results in expressions like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). L'Hôpital's Rule allows you to differentiate the numerator and the denominator until the limit can be directly evaluated without indeterminacies.

• To apply L'Hôpital's Rule, ensure the limit initially results in an indeterminate form.
• Differentiate the top and bottom separately.
• Retry plugging in the value into the newly derived ratio.

For instance, if we have \(\lim_{x \to 0} \frac{\sin(x)}{x}\), directly substituting \(0\) gives \(\frac{0}{0}\). Here, using L'Hôpital's Rule by differentiating top and bottom results in \(\lim_{x \to 0} \frac{\cos(x)}{1} = 1\). It simplifies complex indeterminate cases and is valuable in understanding the behavior of functions.

Trigonometric Limits

Trigonometric limits are specific limits involving trigonometric functions such as sine, cosine, and secant as the variable approaches a certain value. Understanding and calculating these limits is crucial, especially as it relates to small-angle approximations.For example, one of the fundamental results we use is \(\lim_{x \to 0} \frac{\sin(x)}{x} = 1\). Similarly, as \(x\) approaches \(0\), \(\cos(x)\) approaches \(1\), and these results help simplify many trigonometric limits.In our exercise, we deal with \(\sec(x)\), which is \(\frac{1}{\cos(x)}\). Knowing that as \(x\) tends towards zero, \(\cos(x)\) tends towards 1, it follows that \(\sec(x)\) tends towards 1 as well. Such insights are vital for simplifying expressions, calculating limits, and understanding the behavior of trigonometric functions in calculus.

Calculating Limits

Calculating limits is a fundamental aspect of calculus that involves evaluating the behavior of a function as it approaches a specific value. The key is to understand how the function behaves near the point of interest, not just plugging in values directly.To calculate a limit, consider the basic steps:

• Simplify the function if possible to make the calculation simpler.
• If direct substitution results in an indeterminate form, consider algebraic manipulation, known limits, or techniques like L'Hôpital's Rule.
• Evaluate the limit by substituting new values or using approximations.

For the given problem, we simplify the trigonometric terms first by understanding that \(\sec^2(x)\) transforms to \(\frac{1}{\cos^2(x)}\), then determining the limit as \(x\to 0\). We evaluated that \(\cos(0) = 1\), leading \(\sec^2(x)\) to approach \(1\) as well. Finally, by substituting into our expression, we ultimately find \(\sqrt{8} = 2\sqrt{2}\). Mastery in calculating limits provides a base for deeper understanding in calculus, especially in tackling complex problems.