Evaluate Limit Of Ln(sin X) / Cos X As X Approaches Π/2

#SEO Title: Evaluating Limit of ln(sin x) / cos x | Step-by-Step Solution

Hey guys! Today, we are diving into a fascinating limit problem from the world of calculus. We're going to evaluate the limit of a function that involves trigonometric and logarithmic components. This kind of problem often pops up in advanced calculus courses and is a great exercise in applying L'Hôpital's Rule and understanding the behavior of functions near specific points. So, buckle up, and let's get started!

The Problem

We need to evaluate the following limit:

$$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\ln (\sin x)}{\cos x}$$

This limit looks tricky at first glance, but we will break it down step by step to make it super easy to understand. First, let's address the elephant in the room. Why can't we just plug in $x = \frac{\pi}{2}$ directly? Well, if we try that, we get:

$\ln(\sin(\frac{\pi}{2})) = \ln(1) = 0$

$\cos(\frac{\pi}{2}) = 0$

So, we end up with $\frac{0}{0}$, which is an indeterminate form. This means we can't determine the limit just by direct substitution, and we need to use another method. The most common technique for dealing with indeterminate forms like this is L'Hôpital's Rule. This rule is a game-changer when it comes to solving limits, so let's refresh our memory on how it works.

Understanding Indeterminate Forms and L'Hôpital's Rule

Before we jump into the solution, let's quickly recap indeterminate forms and L'Hôpital's Rule. Indeterminate forms, like $0/0$, $\infty/\infty$, $0 \cdot \infty$, $\infty - \infty$, $1^\infty$, $0^0$, and $\infty^0$, arise when direct substitution leads to an undefined expression. These forms don't automatically mean the limit doesn't exist; they just mean we need a different approach to evaluate it.

L'Hôpital's Rule is our go-to method for handling many indeterminate forms, especially $0/0$ and $\infty/\infty$. It states that if we have a limit of the form:

$$\lim_{x \to c} \frac{f(x)}{g(x)}$$

where both $f(x)$ and $g(x)$ approach 0 or both approach $\pm \infty$ as $x$ approaches $c$, and if $f'(x)$ and $g'(x)$ exist and $g'(x) \neq 0$ near $c$, then:

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

In simpler terms, if we get an indeterminate form of $0/0$ or $\infty/\infty$, we can take the derivative of the numerator and the derivative of the denominator separately and then try evaluating the limit again. This can often simplify the expression and allow us to find the limit.

Applying L'Hôpital's Rule

Okay, now that we've refreshed our understanding of L'Hôpital's Rule, let's apply it to our problem. We have:

$$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\ln (\sin x)}{\cos x}$$

As we saw earlier, this limit results in the indeterminate form $0/0$ when we substitute $x = \frac{\pi}{2}$. So, L'Hôpital's Rule is perfect for this situation.

First, we need to find the derivatives of the numerator and the denominator separately.

Let $f(x) = \ln(\sin x)$. Using the chain rule, the derivative $f'(x)$ is:

$$f'(x) = \frac{1}{\sin x} \cdot \cos x = \frac{\cos x}{\sin x} = \cot x$$

Now, let $g(x) = \cos x$. The derivative $g'(x)$ is:

$$g'(x) = -\sin x$$

Now we apply L'Hôpital's Rule, which involves dividing the derivative of the numerator by the derivative of the denominator:

$$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\ln (\sin x)}{\cos x} = \lim _{x \rightarrow \frac{\pi}{2}} \frac{\cot x}{-\sin x}$$

We've transformed our original limit into a new one. Let's see if we can evaluate it now.

Simplifying and Evaluating the New Limit

We now have:

$$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cot x}{-\sin x}$$

We can rewrite $\cot x$ as $\frac{\cos x}{\sin x}$, so the limit becomes:

$$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\frac{\cos x}{\sin x}}{-\sin x} = \lim _{x \rightarrow \frac{\pi}{2}} \frac{\cos x}{-\sin^2 x}$$

Now, let's try plugging in $x = \frac{\pi}{2}$ again:

$\cos(\frac{\pi}{2}) = 0$

$\sin(\frac{\pi}{2}) = 1$, so $-\sin^2(\frac{\pi}{2}) = -1$

So, we have:

$$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cos x}{-\sin^2 x} = \frac{0}{-1} = 0$$

We've successfully evaluated the limit! The result is 0.

Final Answer

Therefore,

$$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\ln (\sin x)}{\cos x} = 0$$

Key Takeaways and Conclusion

Key takeaways from this exercise include understanding indeterminate forms and applying L'Hôpital's Rule. This rule is a powerful tool for evaluating limits that initially appear undefined. Additionally, knowing your trigonometric derivatives and identities is crucial for simplifying expressions. We were able to successfully navigate this problem by carefully applying L'Hôpital's Rule and simplifying trigonometric functions.

So, there you have it! We've successfully evaluated the limit of $\frac{\ln(\sin x)}{\cos x}$ as $x$ approaches $\frac{\pi}{2}$. This problem is a great example of how combining different calculus techniques can lead to a solution. Keep practicing, and you'll become a limit-solving pro in no time! Hope you guys found this helpful and engaging! Happy calculating!