Problem 22 - Calculus for Biology and Medicine

Question

Find the limits in Problems 1-60; not all limits require use of l'Hôpital's rule. $$ \lim _{x \rightarrow \infty} \frac{x^{7}}{e^{x}} $$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Identify the Indeterminate Form

We first need to determine the form of the limit as \( x \) approaches infinity. As \( x \rightarrow \infty \), \( x^7 \rightarrow \infty \) and \( e^x \rightarrow \infty \). Thus, it is an indeterminate form of type \( \frac{\infty}{\infty} \), suggesting that l'Hôpital's Rule might be applicable.

Step 2: Apply l'Hôpital's Rule

Since the limit is in the \( \frac{\infty}{\infty} \) form, we can apply l'Hôpital's Rule. This rule states that for limits of \( \frac{f(x)}{g(x)} \) of indeterminate forms, \( \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \) if the limit exists.

Step 3: Calculate First Derivatives

The derivative of the numerator \( f(x) = x^7 \) is \( f'(x) = 7x^6 \). The derivative of the denominator \( g(x) = e^x \) is \( g'(x) = e^x \). The limit now becomes: \[ \lim_{x \to \infty} \frac{7x^6}{e^x} \]

Step 4: Reapply l'Hôpital's Rule

The new limit \( \lim_{x \to \infty} \frac{7x^6}{e^x} \) is still of the \( \frac{\infty}{\infty} \) form, so apply l'Hôpital's Rule again. Differentiate the numerator and the denominator again.

Step 5: Calculate Second Derivatives

The derivative of the new numerator \( 7x^6 \) is \( f''(x) = 42x^5 \). The derivative of the denominator \( e^x \) remains \( g''(x) = e^x \). Reapplying gives us:\[ \lim_{x \to \infty} \frac{42x^5}{e^x} \]

Step 6: Continue Applying l'Hôpital's Rule

The limit remains in the \( \frac{\infty}{\infty} \) form, allowing for l'Hôpital's Rule to be applied repeatedly until the numerator becomes a constant. Continue differentiating until reaching a constant term.

Step 7: Final Derivative and Evaluate the Limit

Eventually, repeated application of l'Hôpital's Rule will result in the numerator being reduced to a constant and the denominator remaining \( e^x \), which approaches infinity. For example, differentiating further reaches:\[ \lim_{x \to \infty} \frac{504}{e^x} \]As \( x \rightarrow \infty \), this expression approaches \( 0 \), since the exponential function grows much faster than any polynomial.

Step 8: Conclude the Limit

After applying l'Hôpital's Rule seven times, the polynomial reduces to a constant over the exponentially increasing \( e^x \), confirming:\[ \lim_{x \to \infty} \frac{x^7}{e^x} = 0 \]

Key Concepts

Indeterminate Forms

When calculating limits, particularly as variables approach infinity or specific points, you might encounter expressions like \( \frac{\infty}{\infty} \), \( \frac{0}{0} \), or other undefined forms. These are known as indeterminate forms. They are crucial in calculus because they signal that the direct evaluation of the limit doesn’t provide useful information about the behavior of the function at that point. In these cases, additional techniques are required to find the limit. Recognizing an indeterminate form is the first step to applying methods like l'Hôpital's Rule or algebraic manipulation, allowing us to simplify the expression and find the true limit. Without identifying indeterminate forms, it would be difficult to progress in solving many limits accurately.

l'Hôpital's Rule

l'Hôpital's Rule is an invaluable tool in calculating limits, especially when faced with indeterminate forms like \( \frac{\infty}{\infty} \) and \( \frac{0}{0} \). The rule states that if \( \lim_{x \to c} \frac{f(x)}{g(x)} \) produces an indeterminate form, and if the derivatives \( f'(x) \) and \( g'(x) \) exist and \( g'(x) eq 0 \) near \( c \), then

• \( \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \)

This rule allows you to simplify the problem by differentiating the numerator and the denominator separately, making it easier to evaluate the limit. In the context of the original exercise, applying the rule repeatedly lets us reduce the polynomial in the numerator until it's eventually overtaken by the growth of the exponential function in the denominator. This approach clearly illustrates the massive growth difference between polynomial and exponential functions, typically leading to a limit of zero as demonstrated.

Exponential Functions

Exponential functions, such as \( e^x \), play a significant role in calculus, particularly in limit problems. They are characterized by their rapid growth compared to polynomial functions like \( x^n \). As \( x \to \infty \), the value of \( e^x \) grows exponentially, overshadowing even the highest degree polynomials in terms of speed of growth. In the context of our problem, while \( x^7 \) becomes very large, \( e^x \) grows so much faster that eventually, the quotient \( \frac{x^7}{e^x} \) approaches zero. This demonstrates a critical principle: even though polynomials may increase without bound, they are eventually dominated by exponential functions over infinite intervals. Understanding this behavior is key in solving many limit problems involving exponential expressions.