Question

In Exercises 2.4.2-2.4.40, find the indicated limits. $$ \lim _{x \rightarrow \infty} x^{\sin (1 / x)} $$

Short answer

Answer: The limit of the function as x approaches infinity is 1.

Step-by-step solution

Rewrite the expression using logarithmic property

First, we will rewrite the function using the property of logarithms. We are going to convert the given function into a format which can be easier to work with. Recall that \(a^b = e^{(\ln a^b)}\). So, $$ x^{\sin (1 / x)} = e^{\ln\left(x^{\sin (1 / x)}\right)} $$

Apply the logarithmic property

Now, we can apply the logarithmic property of power \(\ln(a^b) = b \ln a\): \begin{align*} e^{\ln\left(x^{\sin (1 / x)}\right)} &= e^{\sin (1 / x) \cdot \ln x}\\ \end{align*}

Rewrite the limit

Now, rewrite the limit using the new expression: $$ \lim_{x \rightarrow \infty} x^{\sin (1 / x)} = \lim_{x \rightarrow \infty} e^{\sin (1 / x) \cdot \ln x} $$

Apply L'Hôpital's Rule to exponential

To find the limit of the exponential part, we can use L'Hôpital's Rule. L'Hôpital's Rule states that if you have a limit of the form \(\lim_{x \to a} \frac{f(x)}{g(x)}\) where \(\lim_{x \to a} f(x) = \lim_{x \to a} g(x) = 0\) or \(\lim_{x \to a} f(x) = \lim_{x \to a} g(x) = \pm\infty\), then: $$ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} $$ provided that the limit on the right-hand side exists or is equal to \(\pm \infty\). In this case, we will apply L'Hôpital's Rule to the limit of the exponent, as x approaches infinity: $$ \lim_{x \rightarrow \infty} \frac{\sin (1 / x)}{1/ \ln x} $$

Find derivatives of numerator and denominator

Let's find the derivatives of the numerator and denominator: 1. Numerator derivative: \(\frac{d}{dx}\left(\sin (1 / x)\right) = -\frac{1}{x^2} \cdot \cos (1 / x)\) 2. Denominator derivative: \(\frac{d}{dx}\left(\frac{1}{\ln x}\right) = -\frac{1}{x (\ln x)^2}\) Now we can rewrite the limit using L'Hôpital's Rule: $$ \lim_{x \rightarrow \infty} \frac{\sin (1 / x)}{1/ \ln x} = \lim_{x \rightarrow \infty} \frac{-\frac{1}{x^2} \cdot \cos (1 / x)}{-\frac{1}{x (\ln x)^2}} $$

Simplify the expression

Simplify the expression inside the limit: $$ \lim_{x \rightarrow \infty} \frac{-\frac{1}{x^2} \cdot \cos (1 / x)}{-\frac{1}{x (\ln x)^2}} = \lim_{x \rightarrow \infty} \frac{\cos (1 / x)}{x (\ln x)^2} $$

Evaluate the limit of the simplified expression

Now evaluate the limit: $$ \lim_{x \rightarrow \infty} \frac{\cos (1 / x)}{x (\ln x)^2} = 0 $$ As x approaches infinity, the numerator becomes \(\cos(0) = 1\), while the denominator becomes infinitely large. This causes the overall fraction to approach zero. Now, plug this result back into the expression from Step 3:

Final Answer

Finally, evaluating the overall limit: $$ \lim_{x \rightarrow \infty} x^{\sin (1 / x)} = \lim_{x \rightarrow \infty} e^{\sin (1 / x) \cdot \ln x} = e^0 = 1 $$ So, the limit of the function as x approaches infinity is 1.

Key concepts

L'Hôpital's Rule

L'Hôpital's Rule is an invaluable tool in calculus for calculating indeterminate limits, such as \(0/0\) or \(\infty/\infty\). The rule states that for functions \(f(x)\) and \(g(x)\), if the limit of \(f(x)/g(x)\) as \(x\) approaches a certain value is indeterminate, then the limit of their derivatives \(f'(x)/g'(x)\) should give the same result, provided this latter limit exists or equals \(\pm\infty\).

In the context of the original exercise, L'Hôpital's Rule is applied to find the limit of \( \sin(1/x) / (1/\ln x) \) as \(x\) approaches infinity. After finding the derivatives of both the numerator and the denominator, the limit becomes clearer and can be simplified, eventually showing that as \(x\) approaches infinity, the expression tends toward zero.

Exponential Functions

Exponential functions are mathematical expressions of the form \(f(x) = b^x\), where \(b\) is a positive real number, and \(beq1\). These functions grow or decay at a rate proportional to their value, which makes them extremely important in various fields including science, finance, and engineering.

In the given problem, an exponential function arises when the expression \(x^{\sin(1/x)}\) is rewritten as \(e^{\sin(1/x)\cdot\ln x}\), utilizing the property \(a^b = e^{b\ln a}\). This is a crucial step that transforms the original limit into a form that is suitable for analysis with tools like L'Hôpital's Rule. The final limit of the transformed exponential function leads to the conclusion that the value is \(e^0\), which equals 1.

Logarithmic Properties

Logarithms are the inverse operations of exponentiation and have properties that make them particularly useful for solving equations involving exponents. Some key properties include the power rule \(\ln(a^b) = b \ln a\), the product rule \(\ln(ab) = \ln a + \ln b\), and the quotient rule \(\ln(a/b) = \ln a - \ln b\).

In the step-by-step solution to the exercise, the power rule is used to simplify the limit of \(x^{\sin(1/x)}\) by transforming it into \(e^{\sin(1/x)\cdot\ln x}\). This manipulation is not just a mathematical trick; it reflects a deeper understanding of logarithmic properties that enable us to handle complex limits involving exponential growth or decay. Understanding these properties can reveal the behavior of functions as their inputs become very large or very small.