Problem 46 - University Calculus: Early Transcendentals

Question

Use I'Hópital's rule to find the limits. $$\lim _{x \rightarrow \infty} x^{2} e^{-x}$$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Understand the Problem

We need to find the limit of the expression \( \lim_{x \to \infty} x^2 e^{-x} \). As \( x \to \infty \), the expression is an indeterminate form \( \infty/\infty \), making it suitable for I'Hôpital's Rule.

Step 2: Rewrite as a Quotient

Rewrite the limit in a form suitable for I'Hôpital's Rule: \( \lim_{x \to \infty} \frac{x^2}{e^x} \).

Step 3: Apply I'Hôpital's Rule First Time

Differentiate the numerator and the denominator separately: - Derivative of \(x^2\) is \(2x\).- Derivative of \(e^x\) is \(e^x\).Now evaluate the limit \( \lim_{x \to \infty} \frac{2x}{e^x} \). This is still an \( \infty/\infty \) form.

Step 4: Apply I'Hôpital's Rule Second Time

Differentiate the numerator and the denominator again: - Derivative of \(2x\) is \(2\).- Derivative of \(e^x\) is \(e^x\).Now evaluate the limit \( \lim_{x \to \infty} \frac{2}{e^x} \).

Step 5: Evaluate the Final Limit

As \(x \to \infty\), \(e^x\) grows exponentially, so \(\frac{2}{e^x} \rightarrow 0\). Hence, the limit is 0.

Key Concepts

L'Hopital's Rule

L'Hopital's Rule is a powerful calculus tool used to find limits of indeterminate forms like \(0/0\) or \(\infty/\infty\). If direct substitution in a limit problem results in these forms, L'Hopital's Rule comes into play.
It allows the differentiation of the numerator and the denominator separately.
If \( \lim_{x \to c} f(x) \) and \( \lim_{x \to c} g(x) \) both lead to 0/0 or \(\infty/\infty\), then:

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

Differentiating might simplify the expression and remove the indeterminate form.
This step can be repeated if needed, as long as the condition remains \(0/0\) or \(\infty/\infty\) after differentiation.

Limits

Limits help us understand how functions behave as the input approaches a certain value.
They are essential in calculus for defining derivatives and integrals. In this context, we are dealing with limits involving infinity.
When taking a limit as \(x \to \infty\), we need to consider how the function grows. For instance, \(x^2\) grows polynomially, while \(e^x\) grows exponentially.
In the expression \( \lim_{x \to \infty} x^2 e^{-x} \), as \(x\) becomes very large, \(e^{-x}\) becomes very small, which influences the behavior of the whole expression.

• A polynomial divided by an exponential function tends to shrink to zero, given that exponentials grow faster than polynomials.

Exponential Functions

Exponential functions, like \(e^x\), are a special class of mathematical functions where the variable is in the exponent.
These functions grow extremely rapidly as \(x\) increases. The base \(e\) is a mathematical constant approximately equal to 2.718.

• The inverse of \(e^x\) is \(e^{-x}\), which decreases rapidly as \(x\) gets larger.

This rapid increase and decrease makes exponential functions heavily impactful in limit problems.
In our expression \(x^2 e^{-x}\), the \(e^{-x}\) part causes the function to shrink to zero as \(x\) approaches infinity, overpowering the polynomial growth of \(x^2\).
Understanding this behavior is crucial to predicting the behavior of limits involving exponential functions.