Evaluating The Limit Of 3x*e^(1/x) - 3x As X Approaches Infinity

In this article, we will explore the evaluation of the limit of the function $3xe^{\frac{1}{x}} - 3x$ as $x$ approaches infinity. This problem falls under the category of limits in calculus, specifically dealing with indeterminate forms. We will utilize techniques such as algebraic manipulation, substitution, and L'Hôpital's Rule to arrive at the solution. Understanding limits is crucial in calculus as it forms the basis for concepts like derivatives and integrals. This particular limit showcases how exponential and linear functions interact as their input grows unbounded, highlighting the power of calculus in analyzing such behaviors.

Understanding the Problem

Before diving into the solution, let's clearly define the problem we are tackling. We are asked to find the limit of the expression $3xe^{\frac{1}{x}} - 3x$ as $x$ approaches infinity. This can be written mathematically as:

$$\lim_{x \to \infty} (3xe^{\frac{1}{x}} - 3x)$$

At first glance, substituting $x = \infty$ directly leads to an indeterminate form. As $x$ approaches infinity, $\frac{1}{x}$ approaches 0, and $e^{\frac{1}{x}}$ approaches $e^0 = 1$. Thus, the expression looks like $\infty * 1 - \infty$, which is an indeterminate form of the type $\infty - \infty$. This means we cannot directly evaluate the limit and need to employ further techniques to resolve the indeterminacy. The challenge lies in transforming the expression into a form where the limit can be easily computed, often by using algebraic manipulations or applying L'Hôpital's Rule. We aim to rewrite the expression in a way that allows us to see how the terms interact as $x$ becomes very large, and ultimately determine the value the expression converges to.

Initial Attempts and Algebraic Manipulation

Our first step in evaluating the limit is to try and simplify the expression. The presence of the term $3x$ in both parts of the expression suggests factoring it out. This gives us:

$$\lim_{x \to \infty} 3x(e^{\frac{1}{x}} - 1)$$

Now, as $x$ approaches infinity, we have $3x$ approaching infinity and $(e^{\frac{1}{x}} - 1)$ approaching $e^0 - 1 = 1 - 1 = 0$. This transforms our indeterminate form to $\infty * 0$, which is still an indeterminate form but a different type. This form is often easier to handle, especially when considering applying L'Hôpital's Rule. To do so, we need to rewrite the expression as a fraction, either in the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$. A common strategy in such cases is to move one of the terms to the denominator by taking its reciprocal. Let's rewrite the expression as:

$$\lim_{x \to \infty} \frac{e^{\frac{1}{x}} - 1}{\frac{1}{3x}}$$

Now, as $x$ approaches infinity, the numerator $(e^{\frac{1}{x}} - 1)$ approaches 0, and the denominator $\frac{1}{3x}$ also approaches 0. This gives us the indeterminate form $\frac{0}{0}$, which is suitable for applying L'Hôpital's Rule. This algebraic manipulation has been crucial in setting up the problem for the next step, where we will employ a powerful calculus tool to find the limit.

Applying L'Hôpital's Rule

Now that we have the limit in the indeterminate form $\frac{0}{0}$, we can apply L'Hôpital's Rule. This rule states that if the limit of $\frac{f(x)}{g(x)}$ as $x$ approaches a value (including infinity) results in an indeterminate form of type $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then:

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

provided the limit on the right-hand side exists. In our case, $f(x) = e^{\frac{1}{x}} - 1$ and $g(x) = \frac{1}{3x}$. We need to find the derivatives of both functions. First, let's find $f'(x)$:

$$f'(x) = \frac{d}{dx}(e^{\frac{1}{x}} - 1) = e^{\frac{1}{x}} * \frac{d}{dx}(\frac{1}{x}) = e^{\frac{1}{x}} * (-\frac{1}{x^2}) = -\frac{e^{\frac{1}{x}}}{x^2}$$

Next, let's find $g'(x)$:

$$g'(x) = \frac{d}{dx}(\frac{1}{3x}) = \frac{1}{3} * \frac{d}{dx}(x^{-1}) = \frac{1}{3} * (-1 * x^{-2}) = -\frac{1}{3x^2}$$

Now, we can apply L'Hôpital's Rule:

$$\lim_{x \to \infty} \frac{e^{\frac{1}{x}} - 1}{\frac{1}{3x}} = \lim_{x \to \infty} \frac{-\frac{e^{\frac{1}{x}}}{x^2}}{-\frac{1}{3x^2}} = \lim_{x \to \infty} \frac{3e^{\frac{1}{x}}}{1}$$

This simplification makes the limit much easier to evaluate. As $x$ approaches infinity, $\frac{1}{x}$ approaches 0, and $e^{\frac{1}{x}}$ approaches $e^0 = 1$. Thus, the limit becomes:

$$\lim_{x \to \infty} 3e^{\frac{1}{x}} = 3 * 1 = 3$$

Therefore, the limit of the original expression as $x$ approaches infinity is 3. The application of L'Hôpital's Rule was crucial in transforming the indeterminate form into a manageable expression, allowing us to directly compute the limit.

Alternative Approach: Substitution

While L'Hôpital's Rule provided an efficient way to solve the limit, let's explore an alternative approach using substitution. This method can sometimes offer a different perspective and reinforce our understanding of the problem. The key idea behind substitution is to introduce a new variable that simplifies the expression, particularly the exponent in this case. Let's substitute $u = \frac{1}{x}$. As $x$ approaches infinity, $u$ approaches 0. This substitution transforms our original limit into:

$$\lim_{x \to \infty} 3x(e^{\frac{1}{x}} - 1) = \lim_{u \to 0} 3(\frac{1}{u})(e^u - 1) = \lim_{u \to 0} \frac{3(e^u - 1)}{u}$$

Now, we have a new limit expression in terms of $u$ that approaches 0. This form is another classic indeterminate form of type $\frac{0}{0}$, making it suitable for applying L'Hôpital's Rule again, or recognizing a standard limit. Let's consider applying L'Hôpital's Rule to this new expression. We need to find the derivatives of the numerator and the denominator. The derivative of the numerator, $3(e^u - 1)$, with respect to $u$ is $3e^u$. The derivative of the denominator, $u$, with respect to $u$ is 1. Applying L'Hôpital's Rule, we get:

$$\lim_{u \to 0} \frac{3(e^u - 1)}{u} = \lim_{u \to 0} \frac{3e^u}{1}$$

Now, as $u$ approaches 0, $e^u$ approaches $e^0 = 1$. Thus, the limit becomes:

$$\lim_{u \to 0} 3e^u = 3 * 1 = 3$$

Alternatively, we could have recognized the limit $\lim_{u \to 0} \frac{e^u - 1}{u}$ as a standard limit, which is equal to 1. Therefore, our limit becomes:

$$\lim_{u \to 0} \frac{3(e^u - 1)}{u} = 3 * \lim_{u \to 0} \frac{e^u - 1}{u} = 3 * 1 = 3$$

Both approaches, using L'Hôpital's Rule directly and recognizing the standard limit, confirm our previous result. The substitution method not only provided an alternative solution but also highlighted the importance of recognizing standard limits and the flexibility of choosing appropriate techniques to evaluate limits. It also emphasized how changing variables can sometimes simplify a problem and make it more accessible.

Conclusion

In conclusion, we have successfully evaluated the limit of the function $3xe^{\frac{1}{x}} - 3x$ as $x$ approaches infinity. We started by identifying the indeterminate form and then employed algebraic manipulation to rewrite the expression into a form suitable for applying L'Hôpital's Rule. This led us to the solution, which is 3. We further reinforced our understanding by exploring an alternative approach using substitution, which also resulted in the same answer. This problem illustrates the importance of understanding various techniques for evaluating limits, including algebraic manipulation, L'Hôpital's Rule, and substitution. Recognizing standard limits and choosing the most appropriate method are crucial skills in calculus. This exercise not only provided a solution to a specific problem but also deepened our understanding of the behavior of functions as their inputs grow unbounded and the power of calculus in analyzing such behaviors. The combination of analytical techniques and a clear understanding of fundamental concepts allows us to effectively tackle complex limit problems.