Problem 19 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find each limit. Be sure you have an indeterminate form before applying l'Hôpital's Rule. $$ \lim _{x \rightarrow 0}\left(x+e^{x / 3}\right)^{3 / x} $$

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Identify the form of the limit

First, substitute the value of 0 into the expression to check if it creates an indeterminate form. By setting \ \( \lim_{x \to 0}(x+e^{x/3})^{3/x} \), we see \(e^{x/3} \) approaches \(e^{0} = 1\) and \(x\) approaches \(0\), making the base \ \(1 + 0 = 1 \). The expression becomes \(1^{\infty}\), which is an indeterminate form.

Step 2: Apply logarithms to simplify

Given the indeterminate form \(1^{\infty}\), take the natural logarithm of the expression to transform it. Define \(y = (x + e^{x/3})^{3/x}\) and then take \(\ln\) on both sides: \ \(\ln(y) = \frac{3}{x} \cdot \ln(x + e^{x/3})\).

Step 3: Evaluate the limit using l'Hôpital's Rule

Now seek \(\lim_{x\to 0} \ln(y)\). Substitute small values near zero and consider the difference in behavior: \ \(\ln(x + e^{x/3}) = \ln(1 + x)\). This results in \ \(\frac{0}{0}\) when substituting \ \(x = 0\), allowing us to use l'Hôpital's Rule. Differentiate the numerator and the denominator: \ \([-\ln(x+1)']\) gives \(\frac{1}{x+1}\), and \([-x']\) is \(-1\), leading to \ \(\frac{-1}{x+1} / -1 = \frac{1}{x+1}\).

Step 4: Simplify using limits

Apply the limit to the simplified expression: \ \(\lim_{x\to 0} \frac{1}{1} = 1\). This shows \(\lim_{x\to 0}\ln(y) = 0\). Hence, \(\lim_{x\to 0} y = e^{0} = 1\).

Step 5: Conclusion

Convert back from the logarithmic form to the exponential form. The original limit is the exponential of the result: \ \(\lim_{x\to 0}(x+e^{x/3})^{3/x} = e^{\lim_{x\to 0} \ln(y)} = e^{0} = 1\).

Key Concepts

l'Hôpital's Rule

When faced with limits that result in indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), l'Hôpital's Rule is a powerful tool. The rule helps by allowing you to find the limit of a fraction. You do this by taking the derivative of both the numerator and the denominator separately.
However, before applying l'Hôpital's Rule, the expression must be in these specific forms. If not, you cannot use the rule and might need to simplify or manipulate the expression using other methods.
It's important to carefully differentiate each part according to the rules of calculus. By substituting back, you find the limit again which generally turns out to be more manageable.

Indeterminate Forms

In calculus, an indeterminate form is a mathematical expression with a limit that can't be directly evaluated. Common indeterminate forms include \( 0/0 \), \( \infty/\infty \), and \( 1^{\infty} \). These forms often appear when calculating limits and must be transformed or manipulated to find a solution.
The expression \( \lim_{x \to 0} (x+e^{x/3})^{3/x} \) forms \( 1^{\infty} \), which is indeterminate. To tackle this, you can apply transformation techniques, read on to how it is often done with logarithmic transformations.
Recognizing when an expression results in an indeterminate form is crucial for finding limits and ensures you can use tools like l'Hôpital's Rule effectively.

Logarithmic Transformation

Transforming an expression with logarithms is a clever strategy to handle indeterminate forms, especially \( 1^{\infty} \). By applying the natural logarithm, you convert the expression into a manageable form.
For example, if \( y = (x + e^{x/3})^{3/x} \), you take the natural logarithm on both sides, simplifying to \( \ln(y) = \frac{3}{x} \cdot \ln(x + e^{x/3}) \).
This transformation often creates a fraction that fits the requirements of l'Hôpital's Rule and makes it easier to evaluate the limit by differentiation. After solving the logarithmic form, you revert by exponentiating the result to get back to the original variables.
Applying this method helps simplify complex expressions, making limits more accessible. It also introduces the beauty of logarithms in solving intricate calculus problems.