Problem 46 - Essential Calculus

Question

\(\begin{array}{llll}\text { Comparing } & \text { Functions } & \text { In Exercises } & \mathbf{4 5}-\mathbf{5 0}, & \text { use } & \text { L'Hôpital's }\end{array}\) Rule to determine the comparative rates of increase of the functions \(f(x)=x^{m}, \quad g(x)=e^{n x},\) and \(h(x)=(\ln x)^{n}\) where \(n>0, m>0,\) and \(x \rightarrow \infty\). \(\lim _{x \rightarrow \infty} \frac{x^{3}}{e^{2 x}}\)

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Understand the problem statement

The problem asks for the limit as \(x\) approaches infinity of the ratio of the function \(x^{3}\) to the function \(e^{2x}\).

Step 2: Application of L'Hopital's Rule

L'Hopital's Rule states that \(\lim _{x \rightarrow a} \frac{f(x)}{g(x)}= \lim _{x \rightarrow a} \frac{f'(x)}{g'(x)}\) if \(\lim _{x \rightarrow a} f(x) = \lim _{x \rightarrow a} g(x) = 0\) or ±∞. Apply the rule to the limit \(\lim _{x \rightarrow \infty} \frac{x^{3}}{e^{2 x}}\). The derivative of \(x^{3}\) is \(3x^{2}\) and the derivative of \(e^{2x}\) is \(2e^{2x}\). The limit after applying L'Hopital's Rule is \(\lim _{x \rightarrow \infty} \frac{3x^{2}}{2e^{2x}}\).

Step 3: Apply L'Hopital's Rule again

Since the fraction still has the form \(\frac{0}{0}\) or \(\frac{∞}{∞}\) when you take the limit, you apply L'Hopital's Rule again. The derivative of \(3x^{2}\) is \(6x\) and the derivative of \(2e^{2x}\) is \(4e^{2x}\). The limit becomes \(\lim _{x \rightarrow \infty} \frac{6x}{4e^{2x}}\).

Step 4: Apply L'Hopital's Rule one more time

We still have the form \(\frac{0}{0}\) or \(\frac{∞}{∞}\), so apply L'Hopital's rule once more. The derivative of \(6x\) is \(6\) and the derivative of \(4e^{2x}\) is \(8e^{2x}\). The limit becomes \(\lim _{x \rightarrow \infty} \frac{6}{8e^{2x}}\).

Step 5: Calculation of the limit

The limit \(\lim _{x \rightarrow \infty} = \frac{6}{8e^{2x}}\) can now be calculated as it's no longer in the form \(\frac{0}{0}\) or \(\frac{∞}{∞}\). As \(x\) approaches infinity, \(e^{2x}\) also approaches infinity so the limit is 0.

Key Concepts

Limits

Limits help us understand the behavior of functions as they approach a particular point or infinity. When we say \( \lim_{x \to a} f(x) \), we're exploring what happens to \( f(x) \) as \( x \) gets really close to \( a \). In this exercise, the goal is to find the limit as \( x \) approaches infinity for a given function, \( \frac{x^3}{e^{2x}} \). This means we're interested in how the ratio of \( x^3 \) to \( e^{2x} \) behaves as \( x \) becomes very large.
Understanding limits is fundamental because it lays the groundwork for more complex mathematical concepts, like derivatives and integrals. If you can master limits, you're well on your way to understanding calculus.

Derivatives

Derivatives tell us how a function changes; in other words, they give us the rate of change or slope of the function at a particular point. In simple terms, if you think of a curve representing \( f(x) \), then the derivative \( f'(x) \) tells us how steep or flat that curve is at any point. In our original problem, we utilize derivatives to apply L'Hôpital's Rule and solve the limit of \( \frac{x^3}{e^{2x}} \).
Each application of L'Hôpital's Rule requires the computation of derivatives. The first derivative of \( x^3 \) is \( 3x^2 \), and the first derivative of \( e^{2x} \) is \( 2e^{2x} \). Continuing to apply this process until a clear result emerges requires understanding the fundamental techniques derived from calculus.

Exponential Functions

Exponential functions are characterized by their rapid growth and have the form \( e^{nx} \). In our exercise, \( e^{2x} \) acts as a comparison to \( x^3 \). The exponential function grows much faster than a polynomial like \( x^3 \). That's a crucial insight needed to anticipate the limit of \( \frac{x^3}{e^{2x}} \) as \( x \to \infty \).
The base \( e \) (approximately 2.718) is unique because the derivative of \( e^x \) is itself, \( e^x \). This self-similar behavior under differentiation makes exponential functions powerful in modeling phenomena across fields like finance, physics, and biology. Understanding this concept explains why \( e^{2x} \) eventually dominates the denominator as \( x \to \infty \), leading the limit to converge to zero.

Comparative Rates of Increase

Comparing the rates of increase between two functions involves determining which function grows faster as \( x \) approaches a certain value, often infinity. The key task is to assess whether the numerator or denominator in a fraction grows more quickly. In our exercise, determining \( \lim_{x \rightarrow \infty} \frac{x^3}{e^{2x}} \) requires us to compare how quickly \( x^3 \) grows versus \( e^{2x} \).
Through multiple applications of L'Hôpital's Rule, we deduce that the exponential function grows much more quickly than the polynomial function. Thus, as \( x \to \infty \), \( e^{2x} \) increases more rapidly, leading the fraction \( \frac{x^3}{e^{2x}} \) toward zero. This comparison highlights the dominance of exponential growth in such scenarios and underscores why exponential functions overpower polynomials in terms of growth rate.