Problem 29 - A Textbook of Differential Calculus

Question

If \(\lim _{x \rightarrow a} f(x)=1\) and \(\lim _{x \rightarrow a} g(x)=\infty\), then \(\lim _{x \rightarrow a}\\{f(x)\\}^{g(x)}=e^{\lim (f(x)-1) \times g(x)}\). \(\lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^{\frac{\sin x}{x-\sin x}}\) is equal to (a) \(\frac{1}{e}\) (b) \(-\frac{1}{e}\) (c) \(e\) (d) \(-e\)

Step-by-Step Solution

Answer

None of the provided choices (ex for this problem. The solution is 1.

Step 1: Identify the Components

First, identify the expressions for \(f(x)\) and \(g(x)\). In this problem, let's set \(f(x) = \frac{\sin x}{x}\) and \(g(x) = \frac{\sin x}{x-\sin x}\).

Step 2: Calculate \(\lim_{x \to 0} f(x)\)

Evaluate the limit \(\lim_{x \to 0} \frac{\sin x}{x}\). This is a standard limit and equals 1. Therefore, \(\lim_{x \to 0} f(x) = 1\).

Step 3: Calculate \(\lim_{x \to 0} g(x)\)

Now compute the limit \(\lim_{x \to 0} \frac{\sin x}{x-\sin x}\). As \(x\) approaches zero, \(\sin x \approx x\), making the denominator small and positive, while the numerator approaches zero. Thus, \(g(x)\) approaches infinity as \(x\) approaches zero.

Step 4: Verify the Limit Condition

Our scenario now aligns with the given condition \(\lim_{x \to a}f(x)=1\) and \(\lim_{x \to a}g(x)=\infty\). We can apply: \(\lim_{x \to a}\{f(x)\}^{g(x)}=e^{\lim (f(x)-1) \times g(x)}\).

Step 5: Calculate \((f(x)-1)\) and Evaluate the Associated Limit

Since \(f(x) = \frac{\sin x}{x}\), then \(f(x) - 1 = \frac{\sin x}{x} - 1 = \frac{\sin x - x}{x}\). Evaluate \(\lim_{x \to 0} \frac{\sin x - x}{x}(\frac{\sin x}{x-\sin x})\) using \(\lim_{x \to 0}\frac{\sin x - x}{x} = 0\). Hence, \(\lim (f(x)-1) \times g(x) = 0\).

Step 6: Substitute Back to Find the Expression Value

Substituting back, we get \(e^{\lim (f(x)-1) \times g(x)} = e^{0} = 1\). This value is consistent with none of the provided answer choices.

Key Concepts

L'Hôpital's Rule

When dealing with limits that result in indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), L'Hôpital's Rule becomes incredibly useful. It provides a way to evaluate these tricky limits by

• Taking the derivative of the numerator and the denominator separately.
• Then, substituting these derivatives back into the limit.

This rule simplifies the process and often reveals the actual limit of the function. Be mindful that it can only be applied if:

• Both numerator and denominator approach 0 or \( \infty \).
• The derivatives exist in a neighborhood of the point.]

This method doesn't solve all indeterminate form problems, but it helps in many cases. Always remember to check if the result is still an indeterminate form after applying it once, and if so, apply it again. Repeat until a definite value is found.

Exponential Functions

Exponential functions are characterized by having a constant base raised to a variable exponent, such as \( f(x) = a^x \), where \( a \) is a positive constant.

• They display rapid growth or decay, depending on whether the base is greater or less than 1.
• In calculus, they can describe continuous processes like compound interest or population growth.
• When extended to real, complex, and negative exponents, their behavior exhibits exponential change regardless of x-values.

A fundamental property is that the function \( e^x \) is its own derivative, which means itself characterizes its rate of change. This unique trait makes it especially convenient in solving calculus problems. Furthermore, these functions also help in approximations near zero, as seen when evaluating limits. When indeterminate forms like \( 1^\infty \) arise, you use the characteristic limit formula to deal with such peculiar situations, often leading to a simple transformation reducing complexity.

Indeterminate Forms

Indeterminate forms occur when the result of direct substitution in limit problems yields an undefined expression such as \( 0/0 \) or \( \infty - \infty \). There are several types of indeterminate forms, including:

• \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \)
• \( 0 \cdot \infty \), \( \infty - \infty \)
• \( 1^\infty \), \( 0^0 \), \( \infty^0 \)

These expressions don't convey enough information about the limit itself without further manipulation.
Using tools like algebraic manipulation or calculus techniques can convert them to a determinate form. L'Hôpital's Rule is particularly useful for the ratios like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), often turning an otherwise complex problem into a solvable one. For exponential indeterminate forms like \( 1^\infty \), often, a logarithmic approach or a related identity for exponential limits simplifies the expression significantly. Identifying and resolving indeterminate forms is key to deepening your understanding in calculus.