Question

Evaluate the following limits. $$\lim _{t \rightarrow \infty} \frac{\cos t}{e^{3 t}}$$

Short answer

Answer: The limit of the function $\frac{\cos t}{e^{3t}}$ as $t$ approaches infinity is 0.

Step-by-step solution

1. Identify the limit

We need to evaluate the limit: $$\lim _{t \rightarrow \infty} \frac{\cos t}{e^{3 t}}$$

2. Analyze the function

As t approaches infinity, the denominator \(e^{3t}\) becomes infinite, while the numerator oscillates between -1 and 1 because it's a cosine function. We can rewrite the limit as: $$\lim _{t \rightarrow \infty} \frac{1}{\frac{e^{3t}}{\cos t}}$$ Now it looks like a limit of the form \(\frac{1}{\infty}\).

3. Apply L'Hopital's Rule

Since the limit has the indeterminate form \(\frac{1}{\infty}\), we can use L'Hopital's rule. We differentiate both the numerator and the denominator with respect to t: $$\lim _{t \rightarrow \infty} \frac{-\sin t}{3e^{3 t}}$$

4. Re-analyze the function

Now the numerator is a sine function oscillating between -1 and 1 as t approaches infinity, and the denominator is still growing infinitely. This still has the indeterminate form of \(\frac{1}{\infty}\).

5. Applying the Squeeze Theorem

As the denominator \(3e^{3t}\) grows infinitely, the value of the overall function will approach zero since the sine function remains bounded between -1 and 1. Therefore, by the Squeeze Theorem: $$\lim _{t \rightarrow \infty} \frac{-\sin t}{3e^{3 t}} = 0$$

6. Conclusion

The limit of the original function is also zero, so: $$\lim _{t \rightarrow \infty} \frac{\cos t}{e^{3 t}} = 0$$

Key concepts

Limits

Limits are a fundamental aspect of calculus, allowing us to evaluate the behavior of a function as it approaches a certain point. In our exercise, we evaluate the limit of a function where the variable approaches infinity. This concept helps us understand how functions behave at extreme values, which is crucial in physics, engineering, and mathematics.

To comprehend limits, think of them as the value that a function "settles into" as the input becomes very large or very small. For example, as the variable "t" approaches infinity in \( \lim_{t \to \infty} \frac{\cos t}{e^{3t}} \), we analyze how the function behaves overall.

• Numerators like \(\cos t\) remain bounded between -1 and 1 as \(t\) goes to infinity.
• The denominator, \(e^{3t}\), grows exponentially, becoming exceedingly large.

The balance between these two gives us insight into where the function heads, leading us to understand why the given limit approaches zero.

L'Hopital's Rule

L'Hopital's Rule is a useful method for solving limits that result in indeterminate forms, specifically \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). When directly substituting into a limit doesn't provide a clear answer, this rule allows the differentiation of both the numerator and denominator to find the limit more easily.

In our exercise, we initially encounter an indeterminate form when evaluating the limit \(\lim _{t \rightarrow \infty} \frac{\cos t}{e^{3 t}}\). While not a classic 0/0 or \(\infty/\infty\) form, the transforming into \(\frac{-\sin t}{3e^{3 t}}\) through derivatives shows the oscillating nature of the functions involved.

• Using L'Hopital's Rule helps simplify problems by breaking them into more manageable parts.
• After taking derivatives, check if the limit can be evaluated directly or requires further simplification.

L'Hopital's Rule is powerful, but it needs careful application. In scenarios where oscillation continues to complicate evaluation, sometimes other methods like the Squeeze Theorem provide more straightforward solutions.

Squeeze Theorem

The Squeeze Theorem is perfect for finding the limits of functions trapped between two other functions with known limits. It is particularly useful when evaluating limits of oscillating functions like sine or cosine that might not stabilize at first glance.

Consider our exercise where the numerically indeterminate limit \(\lim _{t \rightarrow \infty} \frac{\cos t}{e^{3 t}}\) was found by re-evaluating it as \(\frac{-\sin t}{3e^{3 t}}\). The challenge is balancing oscillation within boundaries as the denominator explodes.

• Because \(-1 \leq -\sin t \leq 1\), and \(e^{3t}\) drives the denominator higher, our function is compressed towards zero.
• Boundary values of sine being constant reinforce that longer "squeeze" leads to a limit of zero as \(t\) becomes very large.

The Squeeze Theorem ensures that if the oscillating component of the function sits tightly bounded, the function limit is as constrained, driving steadily to the boundary limit as well.

Exponential Functions

Exponential functions, indicated by \(e^{kt}\) where \(k > 0\), grow rapidly as their exponent increases. They play crucial roles across science and mathematics, notably in calculating growth and decay processes. In our problem, the function \(e^{3t}\) is part of a denominator, showcasing its dynamic: driving values towards zero as t grows.

Key properties of exponential functions like \(e^{3t}\) include:

• They grow faster than polynomial functions.
• Handle large inputs energetically, representing extreme growth or decay scenarios.

In the limit context, an exponential in the denominator ensures functions with bounded numerators trend to zero. Thus, exponential behavior defines how dominant characteristics of the function emerge when determining limits like \(\lim_{t \to \infty} \frac{\cos t}{e^{3t}}\). By overwhelming lesser fluctuating parts of equations, they lead to concrete simplifying trends crucial for evaluating limits effectively.