Evaluating The Limit Of (sin(3x) - 4x) / (e^(7x) - X^5 - 1)

Introduction

In this article, we will delve into the evaluation of the limit: $\lim_{x \rightarrow 0} \frac{\sin(3x) - 4x}{e^{7x} - x^5 - 1}$. This is a classic calculus problem that requires a solid understanding of limits, L'Hôpital's Rule, and Taylor series expansions. The presence of trigonometric, exponential, and polynomial terms in the expression makes it an interesting and challenging exercise. We will explore different approaches to solve this problem, emphasizing the importance of choosing the right technique for efficient and accurate evaluation. Understanding such limits is crucial in various fields like physics, engineering, and economics, where continuous changes and approximations are frequently analyzed.

Understanding the Limit

Before diving into the solution, it is essential to understand what a limit represents. In calculus, a limit is the value that a function approaches as the input (or independent variable) approaches a certain value. In our case, we are interested in the behavior of the function $f(x) = \frac{\sin(3x) - 4x}{e^{7x} - x^5 - 1}$ as $x$ gets closer and closer to 0. The limit exists if the function approaches a specific value from both the left and the right sides of the point in question. However, merely substituting $x = 0$ into the function yields an indeterminate form, $\frac{0}{0}$, which does not provide the limit's value directly. This is where the power of L'Hôpital's Rule and Taylor series expansions comes into play. These techniques allow us to handle such indeterminate forms and find the true limit of the function. By mastering the concept of limits, you gain a fundamental tool for understanding continuity, derivatives, and integrals, which are the building blocks of calculus and its applications.

Methods to Evaluate the Limit

There are primarily two effective methods to evaluate this limit: L'Hôpital's Rule and Taylor series expansions. L'Hôpital's Rule is a powerful technique that allows us to evaluate limits of indeterminate forms such as $\frac{0}{0}$ or $\frac{\infty}{\infty}$. It states that if the limit of $\frac{f(x)}{g(x)}$ as $x$ approaches a certain value is an indeterminate form, then the limit is equal to the limit of $\frac{f'(x)}{g'(x)}$, provided the latter limit exists. This rule can be applied repeatedly until the limit can be evaluated directly. On the other hand, Taylor series expansions provide a way to approximate functions using an infinite sum of terms based on the function's derivatives at a single point. For functions like $\sin(x)$ and $e^x$, the Taylor series expansions are well-known and can be used to simplify the expression inside the limit. By replacing the functions with their Taylor series approximations, we can often eliminate the indeterminate form and evaluate the limit more easily. Both methods have their advantages and disadvantages, and the choice of method often depends on the specific problem at hand. In this case, we will demonstrate both approaches to provide a comprehensive understanding of the solution.

Method 1: L'Hôpital's Rule

L'Hôpital's Rule is particularly useful when dealing with indeterminate forms, such as $\frac{0}{0}$ or $\frac{\infty}{\infty}$. In our case, as $x$ approaches 0, both the numerator, $\sin(3x) - 4x$, and the denominator, $e^{7x} - x^5 - 1$, approach 0. This confirms that we have an indeterminate form of type $\frac{0}{0}$, making L'Hôpital's Rule applicable. The rule states that if $\lim_{x \rightarrow c} \frac{f(x)}{g(x)}$ 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. Therefore, we need to differentiate both the numerator and the denominator with respect to $x$. The derivative of the numerator, $\sin(3x) - 4x$, is $3\cos(3x) - 4$, and the derivative of the denominator, $e^{7x} - x^5 - 1$, is $7e^{7x} - 5x^4$. Now we have a new limit to evaluate: $\lim_{x \rightarrow 0} \frac{3\cos(3x) - 4}{7e^{7x} - 5x^4}$. If we substitute $x = 0$ into this new expression, we get $\frac{3\cos(0) - 4}{7e^{0} - 5(0)^4} = \frac{3 - 4}{7 - 0} = \frac{-1}{7}$. Since this is a definite value, we have successfully found the limit using L'Hôpital's Rule. This illustrates the power of L'Hôpital's Rule in simplifying complex limits by reducing them to simpler forms through differentiation.

First Application of L'Hôpital's Rule

Applying L'Hôpital's Rule for the first time involves differentiating both the numerator and the denominator of our original function. The numerator is given by $f(x) = \sin(3x) - 4x$, and its derivative, $f'(x)$, is found using the chain rule and basic differentiation rules. The derivative of $\sin(3x)$ is $3\cos(3x)$, and the derivative of $-4x$ is $-4$. Thus, $f'(x) = 3\cos(3x) - 4$. For the denominator, $g(x) = e^{7x} - x^5 - 1$, its derivative, $g'(x)$, is calculated similarly. The derivative of $e^{7x}$ is $7e^{7x}$, and the derivative of $-x^5$ is $-5x^4$. The derivative of the constant $-1$ is 0. Therefore, $g'(x) = 7e^{7x} - 5x^4$. Now, we form the new fraction $\frac{f'(x)}{g'(x)} = \frac{3\cos(3x) - 4}{7e^{7x} - 5x^4}$. We need to evaluate the limit of this fraction as $x$ approaches 0. Substituting $x = 0$ into this expression, we get $\frac{3\cos(0) - 4}{7e^{0} - 5(0)^4} = \frac{3(1) - 4}{7(1) - 0} = \frac{-1}{7}$. This direct substitution gives us a finite value, indicating that we have successfully found the limit after the first application of L'Hôpital's Rule. This step is crucial as it transforms the original indeterminate form into a determinate one, allowing us to directly compute the limit. The careful application of differentiation rules is essential in this process to ensure the correct derivatives are obtained.

Evaluating the Resulting Limit

After applying L'Hôpital's Rule once, we obtained the new limit: $\lim_{x \rightarrow 0} \frac{3\cos(3x) - 4}{7e^{7x} - 5x^4}$. To evaluate this limit, we substitute $x = 0$ into the expression. This gives us $\frac{3\cos(3(0)) - 4}{7e^{7(0)} - 5(0)^4}$. Simplifying this, we have $\frac{3\cos(0) - 4}{7e^{0} - 0}$. Since $\cos(0) = 1$ and $e^{0} = 1$, the expression becomes $\frac{3(1) - 4}{7(1) - 0} = \frac{3 - 4}{7} = \frac{-1}{7}$. This result shows that the limit of the original function as $x$ approaches 0 is $-\frac{1}{7}$. The process of evaluating the resulting limit involves careful substitution and simplification. The fact that we obtained a finite value confirms that our application of L'Hôpital's Rule was successful in resolving the indeterminate form. The resulting limit provides valuable information about the behavior of the function near the point $x = 0$. This method demonstrates how L'Hôpital's Rule can be a powerful tool for solving complex limit problems, especially those involving indeterminate forms.

Method 2: Taylor Series Expansion

Another powerful technique for evaluating limits, particularly those involving trigonometric and exponential functions, is the Taylor series expansion. A Taylor series represents a function as an infinite sum of terms, each involving a derivative of the function evaluated at a specific point. For our problem, we will use the Taylor series expansions of $\sin(3x)$ and $e^{7x}$ around $x = 0$. The Taylor series for $\sin(x)$ around $x = 0$ is given by $x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$, so the Taylor series for $\sin(3x)$ is $3x - \frac{(3x)^3}{3!} + \frac{(3x)^5}{5!} - \cdots = 3x - \frac{9x^3}{2} + \frac{81x^5}{40} - \cdots$. Similarly, the Taylor series for $e^x$ around $x = 0$ is $1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$, so the Taylor series for $e^{7x}$ is $1 + 7x + \frac{(7x)^2}{2!} + \frac{(7x)^3}{3!} + \cdots = 1 + 7x + \frac{49x^2}{2} + \frac{343x^3}{6} + \cdots$. Now, we substitute these expansions into our limit expression. The numerator becomes $(3x - \frac{9x^3}{2} + \cdots) - 4x = -x - \frac{9x^3}{2} + \cdots$, and the denominator becomes $(1 + 7x + \frac{49x^2}{2} + \frac{343x^3}{6} + \cdots) - x^5 - 1 = 7x + \frac{49x^2}{2} + \frac{343x^3}{6} + \cdots - x^5$. We can now rewrite the limit as $\lim_{x \rightarrow 0} \frac{-x - \frac{9x^3}{2} + \cdots}{7x + \frac{49x^2}{2} + \frac{343x^3}{6} + \cdots - x^5}$. Dividing both the numerator and the denominator by $x$, we get $\lim_{x \rightarrow 0} \frac{-1 - \frac{9x^2}{2} + \cdots}{7 + \frac{49x}{2} + \frac{343x^2}{6} + \cdots - x^4}$. As $x$ approaches 0, all terms with $x$ in them approach 0, so the limit becomes $\frac{-1}{7}$. This result matches the one we obtained using L'Hôpital's Rule, demonstrating the consistency and effectiveness of the Taylor series method.

Expanding sin(3x) and e^(7x) using Taylor Series

To apply the Taylor series method, we first need to find the Taylor series expansions of $\sin(3x)$ and $e^{7x}$ around $x = 0$. The Taylor series expansion of a function $f(x)$ around $x = 0$ is given by $f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots$. For $\sin(3x)$, we have $f(x) = \sin(3x)$. The derivatives are: $f'(x) = 3\cos(3x)$, $f''(x) = -9\sin(3x)$, $f'''(x) = -27\cos(3x)$, and so on. Evaluating these at $x = 0$, we get $f(0) = \sin(0) = 0$, $f'(0) = 3\cos(0) = 3$, $f''(0) = -9\sin(0) = 0$, $f'''(0) = -27\cos(0) = -27$. Thus, the Taylor series for $\sin(3x)$ is $3x - \frac{27x^3}{3!} + \cdots = 3x - \frac{9x^3}{2} + \cdots$. For $e^{7x}$, we have $g(x) = e^{7x}$. The derivatives are: $g'(x) = 7e^{7x}$, $g''(x) = 49e^{7x}$, $g'''(x) = 343e^{7x}$, and so on. Evaluating these at $x = 0$, we get $g(0) = e^{0} = 1$, $g'(0) = 7e^{0} = 7$, $g''(0) = 49e^{0} = 49$, $g'''(0) = 343e^{0} = 343$. Thus, the Taylor series for $e^{7x}$ is $1 + 7x + \frac{49x^2}{2!} + \frac{343x^3}{3!} + \cdots = 1 + 7x + \frac{49x^2}{2} + \frac{343x^3}{6} + \cdots$. These expansions are crucial for simplifying the limit expression and eliminating the indeterminate form. The careful calculation of derivatives and their evaluation at $x = 0$ are essential steps in this process. The Taylor series expansions provide a local approximation of the functions, which is particularly useful for evaluating limits as $x$ approaches a specific value.

Substituting the Expansions and Simplifying

After obtaining the Taylor series expansions for $\sin(3x)$ and $e^{7x}$, we substitute these into the original limit expression. The Taylor series for $\sin(3x)$ is $3x - \frac{9x^3}{2} + O(x^5)$, where $O(x^5)$ represents terms of order $x^5$ and higher. The Taylor series for $e^{7x}$ is $1 + 7x + \frac{49x^2}{2} + \frac{343x^3}{6} + O(x^4)$. Substituting these into the numerator, $\sin(3x) - 4x$, we get $(3x - \frac{9x^3}{2} + O(x^5)) - 4x = -x - \frac{9x^3}{2} + O(x^5)$. For the denominator, $e^{7x} - x^5 - 1$, we get $(1 + 7x + \frac{49x^2}{2} + \frac{343x^3}{6} + O(x^4)) - x^5 - 1 = 7x + \frac{49x^2}{2} + \frac{343x^3}{6} + O(x^4) - x^5$. Now, we rewrite the limit as $\lim_{x \rightarrow 0} \frac{-x - \frac{9x^3}{2} + O(x^5)}{7x + \frac{49x^2}{2} + \frac{343x^3}{6} + O(x^4) - x^5}$. To simplify this, we divide both the numerator and the denominator by $x$, which gives us $\lim_{x \rightarrow 0} \frac{-1 - \frac{9x^2}{2} + O(x^4)}{7 + \frac{49x}{2} + \frac{343x^2}{6} + O(x^3) - x^4}$. This simplification is crucial because it allows us to eliminate the indeterminate form by canceling out the common factor of $x$. The process of substituting the expansions and simplifying the expression sets the stage for evaluating the limit by direct substitution.

Final Evaluation Using Taylor Series

After substituting the Taylor series expansions and simplifying the expression, we have the limit: $\lim_{x \rightarrow 0} \frac{-1 - \frac{9x^2}{2} + O(x^4)}{7 + \frac{49x}{2} + \frac{343x^2}{6} + O(x^3) - x^4}$. To evaluate this limit, we substitute $x = 0$ into the expression. As $x$ approaches 0, all terms containing $x$ will approach 0. Thus, the numerator approaches $-1$, and the denominator approaches $7$. Therefore, the limit is $\frac{-1}{7}$. This result is consistent with the result we obtained using L'Hôpital's Rule, which confirms the correctness of our calculations. The final evaluation step demonstrates the power of Taylor series expansions in simplifying complex limits. By approximating functions with their Taylor series, we can often transform an indeterminate form into a determinate one, allowing for direct evaluation. This method is particularly useful for functions with well-known Taylor series, such as trigonometric and exponential functions. The consistency between the results obtained from L'Hôpital's Rule and Taylor series expansions further reinforces the validity of both methods.

Conclusion

In conclusion, we have successfully evaluated the limit $\lim_{x \rightarrow 0} \frac{\sin(3x) - 4x}{e^{7x} - x^5 - 1}$ using two different methods: L'Hôpital's Rule and Taylor series expansions. Both methods yielded the same result, $-\frac{1}{7}$, demonstrating the consistency and reliability of these techniques in calculus. L'Hôpital's Rule involves repeatedly differentiating the numerator and the denominator until the limit can be evaluated directly, which is particularly useful for indeterminate forms. On the other hand, Taylor series expansions approximate functions using infinite sums, allowing for simplification of the expression and subsequent evaluation of the limit. The choice of method often depends on the specific problem, but understanding both approaches provides a comprehensive toolkit for solving limit problems. Mastering these techniques is crucial for further studies in calculus and its applications in various scientific and engineering fields. The ability to evaluate limits accurately is fundamental to understanding concepts such as continuity, derivatives, and integrals, which are the building blocks of advanced mathematical analysis.