Evaluating The Limit Of (1 + Sin(3x))^(5/x) As X Approaches 0+

Introduction

In the realm of calculus, evaluating limits stands as a fundamental concept, pivotal for understanding the behavior of functions. This article delves into the intriguing limit of the function $(1 + \sin(3x))^{5/x}$ as $x$ approaches $0$ from the positive side, denoted as $\lim_{x \rightarrow 0^{+}}(1 + \sin(3x))^{5/x}$. This particular limit exemplifies the complexities often encountered when dealing with indeterminate forms, specifically of the type $1^\infty$. To unravel this mathematical puzzle, we will employ a combination of techniques, including logarithmic transformation, L'Hôpital's Rule, and a careful examination of the properties of trigonometric functions. This exploration not only provides a solution to the limit but also illuminates the broader principles of limit evaluation and their applications in various mathematical contexts.

Understanding the Indeterminate Form

The expression $(1 + \sin(3x))^{5/x}$ presents an indeterminate form as $x$ approaches $0^{+}$. The term inside the parenthesis, $(1 + \sin(3x))$, approaches 1 since $\sin(3x)$ tends to 0 as $x$ approaches 0. Meanwhile, the exponent $5/x$ grows infinitely large. This situation, characterized as $1^\infty$, does not have an immediately obvious outcome. It could potentially converge to 1, infinity, or any value in between, necessitating a more rigorous method for evaluation. The indeterminate nature arises because the base approaching 1 and the exponent approaching infinity exert conflicting influences on the overall value of the expression. To resolve this indeterminacy, we need to manipulate the expression into a form where standard limit evaluation techniques, such as L'Hôpital's Rule, can be applied effectively. This typically involves transforming the expression using logarithms, which allows us to convert the exponent into a multiplicative factor, making the limit more tractable.

The Strategy: Logarithmic Transformation

To tackle this indeterminate form, a common strategy involves applying a logarithmic transformation. By taking the natural logarithm of the function, we can transform the exponential expression into a product, making it easier to analyze the limit. Let $y = (1 + \sin(3x))^{5/x}$. Taking the natural logarithm of both sides, we get:

$\ln(y) = \ln((1 + \sin(3x))^{5/x})$

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

$\ln(y) = \frac{5}{x} \ln(1 + \sin(3x))$

Now, we need to find the limit of $\ln(y)$ as $x$ approaches $0^{+}$:

$\lim_{x \rightarrow 0^{+}} \ln(y) = \lim_{x \rightarrow 0^{+}} \frac{5 \ln(1 + \sin(3x))}{x}$

This transformation is crucial because it converts the indeterminate form $1^\infty$ into a more manageable form, specifically $\frac{0}{0}$, which is amenable to L'Hôpital's Rule. The logarithmic transformation effectively brings the exponent down, turning the exponential function into a product of functions, which simplifies the limit evaluation process. By finding the limit of the natural logarithm of the function, we can then exponentiate the result to obtain the original limit.

Applying L'Hôpital's Rule

The limit $\lim_{x \rightarrow 0^{+}} \frac{5 \ln(1 + \sin(3x))}{x}$ now presents itself in the indeterminate form $\frac{0}{0}$, as both the numerator and the denominator approach 0 as $x$ approaches $0^{+}$. This is because $\ln(1 + \sin(3x))$ approaches $\ln(1) = 0$ and $x$ approaches 0. This situation perfectly sets the stage for the application of L'Hôpital's Rule. L'Hôpital's Rule states that if the limit of $\frac{f(x)}{g(x)}$ as $x$ approaches $c$ is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, 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-hand side exists.

Differentiating Numerator and Denominator

To apply L'Hôpital's Rule, we need to differentiate the numerator and the denominator separately. Let $f(x) = 5 \ln(1 + \sin(3x))$ and $g(x) = x$. Then, we find the derivatives $f'(x)$ and $g'(x)$.

First, let's find $f'(x)$:

$f'(x) = \frac{d}{dx} [5 \ln(1 + \sin(3x))]$

Using the chain rule, we have:

$f'(x) = 5 \cdot \frac{1}{1 + \sin(3x)} \cdot \frac{d}{dx} (1 + \sin(3x))$

$f'(x) = 5 \cdot \frac{1}{1 + \sin(3x)} \cdot (3 \cos(3x))$

$f'(x) = \frac{15 \cos(3x)}{1 + \sin(3x)}$

Next, we find $g'(x)$:

$g'(x) = \frac{d}{dx} [x] = 1$

Now that we have $f'(x)$ and $g'(x)$, we can apply L'Hôpital's Rule.

Applying L'Hôpital's Rule

Applying L'Hôpital's Rule, we get:

$\lim_{x \rightarrow 0^{+}} \frac{5 \ln(1 + \sin(3x))}{x} = \lim_{x \rightarrow 0^{+}} \frac{\frac{15 \cos(3x)}{1 + \sin(3x)}}{1}$

$\lim_{x \rightarrow 0^{+}} \frac{15 \cos(3x)}{1 + \sin(3x)}$

Now, we can directly substitute $x = 0$ into the expression:

$\frac{15 \cos(3 \cdot 0)}{1 + \sin(3 \cdot 0)} = \frac{15 \cos(0)}{1 + \sin(0)} = \frac{15 \cdot 1}{1 + 0} = 15$

Thus, we have found that:

$\lim_{x \rightarrow 0^{+}} \ln(y) = 15$

This result is a critical intermediate step. It tells us the limit of the natural logarithm of our original function. To find the limit of the original function, we need to exponentiate this result.

Exponentiating to Find the Original Limit

Having found the limit of the natural logarithm of our function, i.e., $\lim_{x \rightarrow 0^{+}} \ln(y) = 15$, we now need to exponentiate this result to obtain the limit of the original function $y = (1 + \sin(3x))^{5/x}$. Recall that we took the natural logarithm of $y$ earlier, so to reverse this operation, we apply the exponential function.

Reversing the Logarithmic Transformation

Since $\lim_{x \rightarrow 0^{+}} \ln(y) = 15$, we can write:

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

Using the property that the exponential function is continuous, we can move the limit inside the exponential:

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

Substituting the value we found earlier:

$e^{\lim_{x \rightarrow 0^{+}} \ln(y)} = e^{15}$

Therefore, the limit of the original function is:

$\lim_{x \rightarrow 0^{+}} (1 + \sin(3x))^{5/x} = e^{15}$

This result provides the final answer to our limit problem. It demonstrates how the combination of logarithmic transformation and L'Hôpital's Rule allows us to navigate indeterminate forms and find precise limits.

Conclusion

In conclusion, we have successfully evaluated the limit $\lim_{x \rightarrow 0^{+}} (1 + \sin(3x))^{5/x}$ using a combination of logarithmic transformation and L'Hôpital's Rule. The initial indeterminate form of $1^\infty$ was addressed by taking the natural logarithm of the function, which transformed the problem into a $\frac{0}{0}$ indeterminate form. L'Hôpital's Rule was then applied to this transformed limit, allowing us to find the limit of the logarithm of the function. Finally, exponentiating this result gave us the limit of the original function, which is $e^{15}$.

Key Takeaways

This exploration highlights several key concepts in calculus:

1. Indeterminate Forms: Recognizing and handling indeterminate forms such as $1^\infty$ is crucial in limit evaluation.
2. Logarithmic Transformation: This technique is often effective in simplifying exponential expressions and converting indeterminate forms into more manageable ones.
3. L'Hôpital's Rule: A powerful tool for evaluating limits of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, L'Hôpital's Rule involves differentiating the numerator and the denominator.
4. Continuity of Functions: The continuity of the exponential function allows us to move the limit inside the function, simplifying the final step of exponentiation.

By mastering these techniques, one can confidently tackle a wide range of limit problems, gaining a deeper understanding of the behavior of functions and their applications in calculus and beyond. The limit $\lim_{x \rightarrow 0^{+}} (1 + \sin(3x))^{5/x} = e^{15}$ serves as a compelling example of the elegance and power of these methods.