Evaluating Limits: A Step-by-Step Guide For (sin 4t)/(cos 5t)

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Hey guys! Today, we're diving into the fascinating world of limits, specifically tackling the problem of evaluating the limit of (sin 4t)/(cos 5t) as t approaches 0. This is a classic calculus problem that often pops up in introductory courses, and understanding how to solve it is super important for grasping more advanced concepts. So, let's break it down step by step and make sure you've got a solid handle on it.

Understanding Limits: The Foundation

Before we jump into the nitty-gritty of this particular problem, let's take a moment to understand what a limit actually is. In simple terms, a limit tells us what value a function approaches as its input gets closer and closer to a specific value. It's not necessarily the value of the function at that point, but rather the value it's heading towards. Think of it like approaching a destination; the limit is the destination itself, not necessarily the exact spot where you currently are.

Mathematically, we write this as:

lim (x → a) f(x) = L

This reads as "the limit of f(x) as x approaches a is equal to L." Here, L is the limit we're trying to find. To really nail the concept, imagine a graph of the function f(x). As you move along the x-axis closer and closer to the value a, the y-values (i.e., the values of f(x)) get closer and closer to L. This visual representation often helps make the idea of a limit more concrete.

The Problem at Hand: lim (t→0) sin(4t) / cos(5t)

Okay, now let's get back to our specific problem:

lim (t→0) sin(4t) / cos(5t)

Our goal is to find out what happens to the function sin(4t) / cos(5t) as t gets closer and closer to 0. The first thing that might come to mind is simply plugging in t = 0 directly into the function. This is a good starting point, but we need to be careful about potential pitfalls, such as dividing by zero.

Direct Substitution: Our First Attempt

Let's try direct substitution. If we plug in t = 0, we get:

sin(4 * 0) / cos(5 * 0) = sin(0) / cos(0) = 0 / 1 = 0

Hey, that was pretty straightforward! We got a definite value, 0. In this case, direct substitution works perfectly. When direct substitution yields a finite number, that's our limit. However, it's crucial to remember that direct substitution doesn't always work, especially when dealing with trigonometric functions and rational expressions. Sometimes, you'll encounter indeterminate forms like 0/0 or ∞/∞, which require further manipulation.

Why Direct Substitution Works Here

You might be wondering why direct substitution worked so smoothly in this particular problem. The reason is that both the sine and cosine functions are continuous. Continuity is a key concept in calculus; it basically means that the function doesn't have any sudden jumps, breaks, or holes. For continuous functions, the limit as x approaches a point is simply the value of the function at that point.

In our case, sin(4t) and cos(5t) are both continuous functions. Cosine, specifically, is continuous and not zero at t=0, which is why we did not encounter any indeterminate forms. This allowed us to confidently plug in t = 0 and get the correct limit.

When Direct Substitution Fails: Indeterminate Forms

Let's briefly touch upon what happens when direct substitution doesn't work. Imagine we had a different function where plugging in the limit value resulted in 0/0. This is an indeterminate form, and it doesn't tell us anything about the limit. It just means we need to do some more work to figure out what's going on.

Common techniques for dealing with indeterminate forms include:

  • Factoring: If you have a rational function (a fraction with polynomials), try factoring the numerator and denominator to see if you can cancel out any common factors.
  • Rationalizing: If you have square roots in your expression, try rationalizing the numerator or denominator (multiplying by a conjugate).
  • L'Hôpital's Rule: This powerful rule (which we'll not use now since direct subsitution already helped us get the answer) states that if you have an indeterminate form of 0/0 or ∞/∞, you can take the derivative of the numerator and the derivative of the denominator and then try evaluating the limit again.
  • Trigonometric Identities: Sometimes, you can use trigonometric identities to rewrite the function in a form that's easier to evaluate.

Additional Examples for Practice

To really solidify your understanding, let's look at a couple more examples:

Example 1:

lim (x→2) (x^2 - 4) / (x - 2)

If we plug in x = 2 directly, we get 0/0, an indeterminate form. So, let's try factoring the numerator:

lim (x→2) (x^2 - 4) / (x - 2) = lim (x→2) (x + 2)(x - 2) / (x - 2)

Now we can cancel the (x - 2) terms:

lim (x→2) (x + 2) = 2 + 2 = 4

So, the limit is 4.

Example 2:

lim (x→0) sin(x) / x

Direct substitution gives us 0/0. This is a classic limit that often appears in calculus. It's a well-known result that:

lim (x→0) sin(x) / x = 1

This limit can be proven using L'Hôpital's Rule or geometric arguments.

Key Takeaways

Okay, let's recap the main points we've covered:

  • Understanding Limits: Limits tell us what value a function approaches as its input gets closer to a specific value.
  • Direct Substitution: Always try direct substitution first. If it yields a finite number, that's your limit!
  • Indeterminate Forms: If you get 0/0 or ∞/∞, you need to use other techniques like factoring, rationalizing, L'Hôpital's Rule, or trigonometric identities.
  • Continuity: Continuous functions make limit evaluation easier because you can directly substitute the limit value.
  • Practice Makes Perfect: The more you practice, the better you'll become at recognizing different types of limit problems and choosing the appropriate techniques.

Back to Our Original Problem

So, circling back to our original problem:

lim (t→0) sin(4t) / cos(5t) = 0

We successfully evaluated this limit using direct substitution because both sin(4t) and cos(5t) are continuous functions, and cos(5t) is not zero at t = 0. The fact that we got a simple, finite answer reinforces the importance of trying direct substitution first.

Wrapping Up

Guys, I hope this step-by-step guide has helped you understand how to evaluate limits, especially when dealing with trigonometric functions. Remember, practice is key, so keep working through different problems and building your skills. Limits are a fundamental concept in calculus, and mastering them will set you up for success in more advanced topics. Happy calculating!