Question

Use limit methods to determine which of the two given functions grows faster, or state that they have comparable growth rates. $$e^{x^{2}} ; x^{x / 10}$$

Short answer

Answer: The function \(e^{x^{2}}\) grows faster than the function \(x^{x/10}\) as \(x\) increases.

Step-by-step solution

Take the Ratio of Two Functions

Take the ratio of the two functions: \(\frac{e^{x^2}}{x^{x/10}}\) This can be rewritten as: $$\frac{e^{x^2}}{e^{\frac{x}{10} \cdot \ln{x}}}$$

Simplify the Ratio

Simplify the ratio by taking the difference of the exponents: $$e^{x^2 - \frac{x}{10} \cdot \ln{x}}$$

Apply L'Hopital's Rule

As we want to find the limit as \(x\) approaches infinity, we apply L'Hopital's rule to the exponents in the expression above: $$\lim_{x\to\infty} \frac{x^2}{(1/10)x \cdot \ln{x}}$$

Simplify the Fraction

Simplify the fraction by cancelling out the \(x\) terms: $$\lim_{x\to\infty} \frac{10x}{\ln{x}}$$

Apply L'Hopital's Rule Once More

Apply L'Hopital's rule to the above fraction: $$\lim_{x\to\infty} \frac{10}{\frac{1}{x}}$$

Evaluate the Limit

As \(x\) approaches infinity, the above limit becomes: $$\lim_{x\to\infty} \frac{10}{\frac{1}{x}} = 10 \lim_{x\to\infty} x = \infty$$

Determine Growth Comparison

Since the limit is infinite, we conclude that the first function \(e^{x^2}\) grows faster than the second function \(x^{x/10}\) as \(x\) approaches infinity.

Key concepts

L'Hopital's Rule

In calculus, L'Hopital's Rule is a handy tool used to evaluate limits involving indeterminate forms. This rule states that for functions that result in \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) form, you can compute the limit of the ratio of their derivatives. Applying L'Hopital's Rule involves differentiating the numerator and the denominator until you achieve a determinate form. For example, given the limit \(\lim_{x\to\infty} \frac{x^2}{(1/10)x \cdot \ln{x}}\), L'Hopital's Rule allows us to differentiate both parts. After differentiating, you'll apply the limit again until a straightforward outcome emerges. Consider why we use this rule: it simplifies complicated expressions making them easier to solve. This rule was applied multiple times during the problem to assess the growth rates of the functions \(e^{x^2}\) and \(x^{x/10}\).

Exponential Growth

Exponential growth describes how quantities increase at increasingly rapid rates. It occurs when the growth rate of a mathematical function is proportional to the function's current value. In the context of this exercise, the function \(e^{x^2}\) represents exponential growth. The presence of the exponential function \(e^x\), known for its rapid growth, showcases how quickly this function escalates. A fascinating aspect of exponential growth is that it can outpace polynomial growth significantly as \(x\) increases, which is evident when comparing \(e^{x^2}\) to \(x^{x/10}\). As the limit evaluation showed, the exponential growth of \(e^{x^2}\) indeed exceeds the other function as \(x\) approaches infinity.

Function Comparison

Comparing functions involves analyzing which grows faster as \(x\) increases indefinitely. An essential method to achieve this is by evaluating the limit of their ratio when \(x\) approaches infinity. In this exercise, the two functions compared are \(e^{x^2}\) and \(x^{x/10}\). By rewriting the ratio \(\frac{e^{x^2}}{x^{x/10}}\), we create a setup where L'Hopital's Rule can be applied. Simplifying these terms reveals insights into their growth behavior. Through calculus, particularly limit evaluation, one can conclude if one function eventually dominates another. In this case, calculating the limit helped us discern that \(e^{x^2}\) grows faster than \(x^{x/10}\) as \(x\) increases, illustrating a fundamental aspect of exponential versus other types of growth.