Problem 45 - University Calculus: Early Transcendentals

Question

Use I'Hópital's rule to find the limits. $$\lim _{t \rightarrow \infty} \frac{e^{t}+t^{2}}{t^{2}-t}$$

Step-by-Step Solution

Answer

The limit is \(\infty\).

Step 1: Identify the Indeterminate Form

First, we need to determine if the limit results in an indeterminate form to decide if L'Hôpital's Rule is applicable. As \(t \rightarrow \infty\), both \(e^t + t^2\) and \(t^2 - t\) approach infinity. Thus, we have the form \(\frac{\infty}{\infty}\), which is indeterminate.

Step 2: Apply L'Hôpital's Rule

Since the limit results in an \(\frac{\infty}{\infty}\) form, we can apply L'Hôpital's Rule. This involves differentiating the numerator and the denominator separately: \(\frac{d}{dt}(e^t + t^2) = e^t + 2t\) and \(\frac{d}{dt}(t^2 - t) = 2t - 1\).

Step 3: Rewrite the Limit using Derivatives

Rewrite the limit as follows using the derivatives of the numerator and denominator: \[ \lim_{t \rightarrow \infty} \frac{e^t + t^2}{t^2 - t} = \lim_{t \rightarrow \infty} \frac{e^t + 2t}{2t - 1} \]

Step 4: Evaluate the New Limit

Now, re-evaluate the limit of the new expression: \( \lim_{t \rightarrow \infty} \frac{e^t + 2t}{2t - 1} \). As \(t \rightarrow \infty\), \(e^t\) dominates \(2t\) and \(2t - 1\), causing the limit to again be of the form \(\frac{\infty}{\infty}\). Therefore, apply L'Hôpital's Rule again.

Step 5: Apply L'Hôpital's Rule Again

Differentiate the new numerator and denominator again: \(\frac{d}{dt}(e^t + 2t) = e^t + 2\) and \(\frac{d}{dt}(2t - 1) = 2\).

Step 6: Simplify and Evaluate the Final Limit

Rewrite the limit after the second application of L'Hôpital's Rule: \[ \lim_{t \rightarrow \infty} \frac{e^t + 2}{2} = \lim_{t \rightarrow \infty} \frac{1}{2} e^t + 1 = \infty \]

Step 7: Conclusion

The limit diverges to infinity, so the final answer is that the original limit is \(\infty\).

Key Concepts

Indeterminate Forms

When dealing with limits, especially those approaching infinity, we occasionally encounter situations called indeterminate forms. These arise when the direct substitution of values results in expressions like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). Such forms do not immediately provide useful information about the actual limit value.

Take the example \( \lim_{t \rightarrow \infty} \frac{e^t + t^2}{t^2 - t} \). As \( t \) becomes very large, both the numerator \( e^t + t^2 \) and the denominator \( t^2 - t \) trend towards infinity. This creates the \( \frac{\infty}{\infty} \) form, indicating an indeterminate situation.

Indeterminate forms suggest the need for further investigation or manipulation of the expression, often using techniques like L'Hôpital's Rule to properly identify the limit.

Limits at Infinity

Limits at infinity aim to determine the behavior of a function as the input variable grows without bound. Understanding this concept is crucial, as it often helps in testing the dominance of different parts of complex functions.

For functions such as \( \lim_{t \rightarrow \infty} \frac{e^t + t^2}{t^2 - t} \), the variable \( t \) approaches infinity. In these cases, it's essential to pinpoint which term in the numerator and denominator grows faster. The faster-growing term defines the behavior of the function as it stretches towards infinity.

In our problem, the exponential function \( e^t \) grows much more rapidly than both \( t^2 \) and \( t \). Hence, we focus on \( e^t \) for determining the limit at infinity. Recognizing such behavior is critical for applying L'Hôpital's Rule effectively when the limit results in an indeterminate form like \( \frac{\infty}{\infty} \).

Exponential Functions

Exponential functions are a fundamental type of function in mathematics, generally expressed in the form \( f(t) = a^t \), where \( a \) is a positive constant. The most common base is \( e \), Euler's number, approximately 2.718. Exponential functions with base \( e \) are denoted as \( e^t \).

A key property of exponential functions is their rapid rate of growth or decay. For instance, \( e^t \) grows much faster than any polynomial function like \( t^2 \) as \( t \to \infty \). This steep growth is why, in limit problems involving exponentials and polynomials, the exponential part often dictates the behavior of the function as it approaches infinity.

In the limit \( \lim_{t \rightarrow \infty} \frac{e^t + t^2}{t^2 - t} \), it is precisely the growth of \( e^t \) that leads to the final limit being infinite. Exponential dominance in such expressions is a frequent reason for the divergence of limits to infinity.