Problem 74 - Thomas Calculus

Question

L'Hopital's Rule does not help with the limits in Exercises \(67-74 .\) Try it- you just keep on cycling. Find the limits some other way. $$ \lim _{x \rightarrow 0^{+}} \frac{x}{e^{-1 / x}} $$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Identify the Indeterminate Form

First, substitute the value that the variable approaches into the expression to identify if it's an indeterminate form. Here, \[ \lim_{x \rightarrow 0^{+}} \frac{x}{e^{-1/x}}. \]Substitute the limit:- Numerator: As \(x \to 0^{+}\), \(x \to 0\).- Denominator: \(e^{-1/x} \to e^{-\infty} = 0\).This gives \(\frac{0}{0}\), an indeterminate form, indicating L'Hopital's Rule might not help.

Step 2: Exponent Transformation

Rewrite the expression to simplify evaluating the limit:\[\lim_{x \to 0^{+}} \frac{x}{e^{-1/x}} = \lim_{x \to 0^{+}} x \cdot e^{1/x}. \]Focus on simplifying \(e^{1/x}\) as \(x \to 0^{+}\).

Step 3: Substitute \(u = 1/x\)

Let \(u = \frac{1}{x}\), so as \(x \to 0^{+}\), \(u \to +\infty\).The expression transforms to:\[\lim_{u \to \infty} \frac{1/u}{e^{u}} = \lim_{u \to \infty} \frac{1}{u e^{u}}. \]Notice the fraction's denominator grows faster than its numerator as \(u \to \infty\).

Step 4: Analyze Growth Rates

In \(\lim_{u \to \infty} \frac{1}{u e^{u}}, \)\(e^{u}\) is exponential growth, and \(u\) is polynomial.Exponential functions grow faster than polynomial ones, so:\[\lim_{u \to \infty} \frac{1}{u e^{u}} = 0.\]

Step 5: Finalize the Original Limit

Having evaluated \(\lim_{u \to \infty} \frac{1}{u e^{u}} = 0\), convert back to variable \(x\):\[\lim_{x \to 0^{+}} \frac{x}{e^{-1/x}} = 0.\]The limit is reached without using L'Hopital's Rule.

Key Concepts

Indeterminate Forms

When you find yourself dealing with limits that result in expressions like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), you've encountered what are known as indeterminate forms. These forms suggest that the limit isn't straightforward and requires more investigation or a different approach to find.

• The notation \(\frac{0}{0}\) indicates that both the numerator and the denominator approach zero as the variable approaches a certain value. This doesn't give information about the behavior of the overall expression.
• Similarly, \(\frac{\infty}{\infty}\) is another indeterminate form where both parts grow without bound.

L'Hopital's Rule can sometimes be used to resolve these forms by taking derivatives, but as the problem suggests, it's not always helpful. In cases where it keeps cycling without simplifying, alternative methods are essential.

Limits

Limits are a fundamental concept in calculus. They describe the behavior of a function as the input approaches a particular value. Whether the input is approaching zero, infinity, or any finite number, limits help determine the expected output of a function in such scenarios.

• Applying limits is crucial when dealing with indeterminate forms, as it helps us find out what the function value tends towards.
• When evaluating limits, especially the tricky ones, rewriting expressions or making substitutions can clarify what the function is doing.

In this exercise, we manipulated the expression to a form that allowed us to see how the terms behave and establish the limit effectively without relying on L'Hopital's Rule.

Exponential Growth

Exponential growth refers to growth that occurs at a constantly increasing rate. In mathematical terms, a function like \(e^x\) represents exponential growth, where each increment of \(x\) results in the value of the function multiplying by a constant factor.

• Exponential functions grow very rapidly compared to other types of functions, such as polynomials.
• In our solution, \(e^{u}\) was dominant in the limit \(\frac{1}{u e^{u}}\) because it grows so much faster than the polynomial term \(u\).

Understanding this rapid growth helps explain why, as \(u\) becomes very large, the overall expression diminishes to zero, since the numerator cannot keep up with the rapidly expanding denominator.

Polynomial Growth

Polynomial growth is characterized by a function whose greatest exponent dictates the rate of growth. Unlike exponential growth, polynomial growth happens at a slower, more predictable rate.

• For a simple polynomial function like \(u\), increasing \(u\) increases the output at a power-dependant rate.
• In the expression \(\frac{1}{u e^{u}}\), the polynomial growth of \(u\) is easily overshadowed by the exponential growth of \(e^{u}\).

By comparing the growth rates of polynomial and exponential functions, it becomes clear why the limit resolves to zero; the exponential term's rapid growth means that the polynomial term becomes insignificant as \(u\) approaches infinity.