Problem 43 - Calculus for Biology and Medicine

Question

Find the limits in Problems 1-60; not all limits require use of l'Hôpital's rule. $$ \lim _{x \rightarrow \infty} x^{1 / x} $$

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Understand the Problem

We need to find the limit of the function \(x^{1/x}\) as \(x\) approaches infinity. This expression raises \(x\) to the power of the reciprocal of \(x\).

Step 2: Simplify the Expression

Rewrite the expression using the property of exponents: \(x^{1/x} = e^{\ln(x^{1/x})} = e^{(1/x) \ln(x)}\). Our task is now to find \(\lim_{x \to \infty} e^{(1/x) \ln(x)}\).

Step 3: Find the Limit of the Exponent

Inside the exponent, we have \((1/x) \ln(x)\). As \(x\) approaches infinity, we consider the behavior of \((1/x) \ln(x)\). The logarithmic function grows slower than any polynomial, and thus \(\ln(x)/x\) will tend towards 0 as \(x\) approaches infinity.

Step 4: Substitute Limit into Exponential Function

Since \(\ln(x)/x\) approaches 0, the limit becomes \(\lim_{x \to \infty} e^{(\ln(x)/x)} = e^0 = 1\).

Step 5: Write the Conclusion

Therefore, the limit of \(x^{1/x}\) as \(x\) approaches infinity is 1.

Key Concepts

l'Hôpital's rule

Imagine you have a function that results in an indeterminate form, like \(0/0\) or \(\infty/\infty\), when calculating its limit. This is where **l'Hôpital's Rule** comes to the rescue. It provides a way to evaluate limits by differentiating the numerator and the denominator separately.
Here's how it works in a nutshell:

• Identify the indeterminate form.
• Differentiate the numerator and the denominator.
• Re-evaluate the limit with these derivatives.

If repeating the process still yields an indeterminate form, you can apply l'Hôpital's Rule again. Simple yet powerful, it makes solving complex limits less daunting.

Exponents in Calculus

Exponents are not just for algebra; they're crucial in calculus too. When we're looking at a limit of a function like \(x^{1/x}\), we're dealing with an exponent where the base \(x\) changes with the variable. This particular case leads us to consider the behavior of exponents when they involve variables.
In our example, the power \(1/x\) approaches zero as \(x\) grows. But we can transform the situation using logarithms and differentiable functions to make it more manageable:

• Rewrite using natural logarithms to simplify the exponent.
• Observe the behavior of the resulting expression as \(x\) approaches infinity.

This method can turn difficult problems into simpler forms that are easier to solve.

Logarithmic Functions

Logarithms can be a bit daunting at first, but they are incredibly useful tools. When trying to simplify problems involving exponents, logarithmic functions provide a way to transform them into manageable expressions.
The natural logarithm, \(\ln(x)\), is a common tool in calculus. It can convert multiplication into addition, division into subtraction, and powers into products. For example,

• Convert \(x^{1/x}\) into \(e^{(1/x)\ln(x)}\).
• This transformation is key for evaluating the behavior of functions as \(x\) approaches infinity.

Using logarithms helps strip away complex layers, making even hard limits approachable.

Infinite Limits

When we talk about limits, some head towards infinity. Understanding **infinite limits** is about grasping how a function behaves as the variable grows unbounded.
Let's consider our example, \(\lim_{x \rightarrow \infty} x^{1/x}\). As \(x\) becomes very, very large, the behaviors of different parts of the function can vary dramatically:

• \(x\), a fast-growing base, competes with the slow-shrinking exponent \(1/x\).
• Even though \(x\) itself becomes infinite, the exponent \(1/x\) trends towards zero, calming the rapid growth.
• Thus, the function approaches a finite limit, in this case 1.

Grasping infinite limits is about studying this balance, recognizing which part grows or shrinks, and in what manner.