Evaluate Limit: (7x)^(sin(4x)) As X->0+

Hey guys! Let's dive into a fun calculus problem today! We're going to evaluate the limit of $(7x)^{\sin(4x)}$ as $x$ approaches $0$ from the positive side. This is a classic example where we need to use logarithms to simplify the expression and apply L'Hôpital's Rule. So, buckle up, and let's get started!

Understanding the Problem

Before we jump into the solution, it's crucial to understand what we're dealing with. We have a function where both the base ($7x$) and the exponent ($\sin(4x)$) are changing as $x$ approaches $0$. Specifically, $7x$ approaches $0$, and $\sin(4x)$ also approaches $0$. This gives us an indeterminate form of the type $0^0$, which is one of the cases where we can't directly substitute the value of $x$ to find the limit. Indeterminate forms require us to use more advanced techniques, such as L'Hôpital's Rule, after transforming the expression into a suitable form.

When we encounter such indeterminate forms, it's a signal that we need to manipulate the expression algebraically or use logarithmic properties to make it easier to evaluate. In this case, logarithms will be our best friend. By taking the natural logarithm of the function, we can bring the exponent down as a coefficient, which will help us transform the indeterminate form into something we can handle with L'Hôpital's Rule. Remember, L'Hôpital's Rule is applicable when we have indeterminate forms of the type $\frac{0}{0}$ or $\frac{\infty}{\infty}$. So, our goal is to rewrite the expression in one of these forms so that we can apply this powerful rule.

Moreover, approaching $0$ from the positive side (denoted as $x \rightarrow 0^{+}$) means that we are only considering values of $x$ that are slightly greater than $0$. This is important because the function $(7x)^{\sin(4x)}$ might not be defined for negative values of $x$, especially if we consider non-integer exponents. Therefore, the limit from the positive side ensures that we are working with a well-defined function in the neighborhood of $0$. This detail is often crucial in mathematical analysis, as it can affect the existence and value of the limit. Always pay attention to the direction from which you are approaching the limit, as it can make a significant difference in the outcome.

Steps to Evaluate the Limit

Step 1: Introduce the Natural Logarithm

Let's denote our function as $y = (7x)^{\sin(4x)}$. To simplify this, we'll take the natural logarithm (ln) of both sides:

$$\ln(y) = \ln((7x)^{\sin(4x)})$$

Using the property of logarithms that $\ln(a^b) = b \ln(a)$, we can rewrite the equation as:

$$\ln(y) = \sin(4x) \ln(7x)$$

Step 2: Rewrite the Expression

Now, we want to rewrite the expression to get it into a form where we can apply L'Hôpital's Rule. We can rewrite the product as a quotient:

$$\ln(y) = \frac{\ln(7x)}{\csc(4x)}$$

Here, we've used the fact that $\csc(4x) = \frac{1}{\sin(4x)}$. As $x \rightarrow 0^{+}$, we have $\ln(7x) \rightarrow -\infty$ and $\csc(4x) \rightarrow \infty$. Thus, we have an indeterminate form of the type $\frac{-\infty}{\infty}$, which is suitable for L'Hôpital's Rule.

Step 3: Apply L'Hôpital's Rule

L'Hôpital's Rule states that if we have a limit of the form $\lim_{x \rightarrow c} \frac{f(x)}{g(x)}$ where both $f(x)$ and $g(x)$ approach $0$ or $\pm \infty$ as $x$ approaches $c$, then:

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

provided the limit on the right exists. In our case, $f(x) = \ln(7x)$ and $g(x) = \csc(4x)$. Let's find their derivatives:

$$f'(x) = \frac{d}{dx} \ln(7x) = \frac{1}{7x} \cdot 7 = \frac{1}{x}$$

$$g'(x) = \frac{d}{dx} \csc(4x) = -4 \csc(4x) \cot(4x)$$

Now, we apply L'Hôpital's Rule:

$$\lim_{x \rightarrow 0^{+}} \frac{\ln(7x)}{\csc(4x)} = \lim_{x \rightarrow 0^{+}} \frac{\frac{1}{x}}{-4 \csc(4x) \cot(4x)} = \lim_{x \rightarrow 0^{+}} \frac{1}{x} \cdot \frac{-1}{4 \csc(4x) \cot(4x)}$$

Step 4: Simplify the Expression Further

We can rewrite the expression as:

$$\lim_{x \rightarrow 0^{+}} \frac{-\sin(4x) \tan(4x)}{4x}$$

Now, we can use the fact that $\lim_{x \rightarrow 0} \frac{\sin(x)}{x} = 1$ and $\lim_{x \rightarrow 0} \frac{\tan(x)}{x} = 1$. So, we rewrite the expression to use these limits:

$$\lim_{x \rightarrow 0^{+}} \frac{-\sin(4x) \tan(4x)}{4x} = \lim_{x \rightarrow 0^{+}} \frac{-\sin(4x)}{4x} \cdot \tan(4x) = \lim_{x \rightarrow 0^{+}} -\frac{\sin(4x)}{4x} \cdot \lim_{x \rightarrow 0^{+}} \tan(4x)$$

We know that $\lim_{x \rightarrow 0^{+}} \frac{\sin(4x)}{4x} = 1$ and $\lim_{x \rightarrow 0^{+}} \tan(4x) = 0$. Therefore,

$$\lim_{x \rightarrow 0^{+}} -\frac{\sin(4x)}{4x} \cdot \tan(4x) = -1 \cdot 0 = 0$$

So, we have:

$$\lim_{x \rightarrow 0^{+}} \ln(y) = 0$$

Step 5: Solve for the Original Limit

Since we found the limit of $\ln(y)$, we need to find the limit of $y$. We know that if $\lim_{x \rightarrow c} \ln(y) = L$, then $\lim_{x \rightarrow c} y = e^L$. In our case, $L = 0$, so:

$$\lim_{x \rightarrow 0^{+}} y = e^0 = 1$$

Therefore, the limit of the original function is:

$$\lim_{x \rightarrow 0^{+}} (7x)^{\sin(4x)} = 1$$

Conclusion

Alright, guys! We've successfully evaluated the limit $\lim_{x \rightarrow 0^{+}} (7x)^{\sin(4x)}$ and found it to be $1$. This problem required us to use logarithms to transform the indeterminate form, apply L'Hôpital's Rule, and simplify the resulting expression. Remember, when you encounter indeterminate forms, think about using logarithms and L'Hôpital's Rule to make your life easier. Keep practicing, and you'll become a pro at these types of problems!

Key takeaways from this problem:

• Indeterminate forms like $0^0$ require special techniques such as using logarithms and L'Hôpital's Rule.
• Taking the natural logarithm can simplify exponential expressions.
• L'Hôpital's Rule is applicable for indeterminate forms of type $\frac{0}{0}$ or $\frac{\infty}{\infty}$.
• Simplifying trigonometric functions using known limits can help in evaluating complex limits.

Hope this helps, and happy calculating!