Problem 5 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Evaluate the following limits. (a) \(\lim _{x \rightarrow \infty} \frac{e^{-x}}{x^{2}}\) (b) \(\lim _{x \rightarrow \infty} x^{3} e^{-3 x}\)

Step-by-Step Solution

Answer

\(\lim _{x \rightarrow \infty} \frac{e^{-x}}{x^{2}} = 0\) and \(\lim _{x \rightarrow \infty} x^{3} e^{-3 x} = 0\).

Step 1: Solving Limit for part (a)

Start by taking the derivative of the numerator and denominator of the function \( \frac{e^{-x}}{x^{2}}\) separately using the chain rule for derivatives. The derivative of \(e^{-x}\) is \(-e^{-x}\) and the derivative of \(x^{2}\) is \(2x\). Hence, our function becomes \(-\frac{e^{-x}}{2x}\). Apply L'Hopital's rule again for \(-\frac{e^{-x}}{2x}\). The derivative of \(-e^{-x}\) is \(e^{-x}\) and the derivative of \(2x\) is \(2\). Thus the new function is \(-\frac{e^{-x}}{2}\). Now the limit can be evaluated directly. As \(x \rightarrow \infty\), \(e^{-x}\) tends to 0.

Step 2: Answer for part (a)

So, \(\lim _{x \rightarrow \infty} \frac{e^{-x}}{x^{2}} = 0\)

Step 3: Solving Limit for part (b)

Firstly take the derivatives of the numerator and the denominator of the function \(x^{3} e^{-3 x}\), which results in \(3x^{2} e^{-3 x} - 3x^{3} e^{-3 x}\). Apply L'Hopital's rule again for \(3x^{2} e^{-3 x} - 3x^{3} e^{-3 x}\). The derivative is \(6x e^{-3 x} - 9x^{2} e^{-3 x}+9x^{3} e^{-3 x}\). We continue this process. In the end, since \(x^{3}\) grows faster than \(e^{-3 x}\) for \(x \rightarrow \infty\), the entire expression will tend towards 0.

Step 4: Answer for part (b)

So, \(\lim _{x \rightarrow \infty} x^{3} e^{-3 x} = 0\)

Key Concepts

L'Hopital's Rule

Understanding L'Hopital's Rule is crucial when evaluating limits that present an indeterminate form, such as \(0/0\) or \(\infty/\infty\). This rule applies to functions that are differentiable in the neighborhood of a point – or at infinity – and it states that if you have a limit of the form \(\lim_{x \to a} \frac{f(x)}{g(x)}\) and both \(f(a)\) and \(g(a)\) are either 0 or \(\infty\), you can use the derivatives of these functions to find the limit.

Simply put, if \(\lim_{x \to a} \frac{f(x)}{g(x)}\) is indeterminate, you can evaluate \(\lim_{x \to a} \frac{f'(x)}{g'(x)}\) instead. If the result is still indeterminate, you may apply L'Hopital's Rule repeatedly until you arrive at a determinate form. In the exercise given, L'Hopital's Rule is applied to evaluate \(\lim_{x \to \infty} \frac{e^{-x}}{x^{2}}\), resulting in an answer of 0 after successive applications of the rule and taking derivatives of the numerator and denominator separately.

Infinite Limits

When evaluating limits that head towards infinity, certain rules are different than those for finite limits. An infinite limit occurs when the value of a function grows without bound as the independent variable approaches either a finite value or infinity itself.

In the context of the given exercise, when \(x\) tends towards infinity, the function's behavior is dependent on how fast the terms \(e^{-x}\) and \(x^2\) or \(x^3 e^{-3x}\) grow or decay. If the growth rate of the denominator is faster than the numerator's, as is the case with \(x^2\) or \(x^3\) in comparison to \(e^{-x}\), the limit of the function will tend towards zero. This comparison of growth rates is integral in analyzing limits that extend to infinity and helps us solve complex problems efficiently.

Exponential Functions

Exponential functions, such as \(e^x\), have unique properties that make them behave differently than polynomial functions, especially as \(x\) approaches infinity. One key property is that an exponential function with a negative exponent – like \(e^{-x}\) – will decay towards zero as \(x\) becomes larger.

This property is pivotal in determining the behavior of functions like those in the textbook exercise. In the case of \(\frac{e^{-x}}{x^{2}}\), the exponential term \(e^{-x}\) decays faster than the polynomial term \(x^2\) grows, resulting in the whole function's limit being zero at infinity. Exponential decay and growth are foundations of not just evaluating limits, but also in fields like calculus, differential equations, and population modeling.