Limit Of ((x-4)/(x+1))^(x+3) As X→∞: Step-by-Step Solution

In this article, we'll dive deep into calculating the limit of the function ((x-4)/(x+1))^(x+3) as x approaches infinity. This problem falls under the categories of limits, functions, exponential functions, and limits without L'Hôpital's rule. Understanding how to solve such limits is crucial in calculus and mathematical analysis. We will break down the solution step by step, making it easy to follow and comprehend.

Understanding the Problem

The problem at hand is to find the limit:

$$\lim_{x\to \infty} \left( \frac{x-4}{x+1} \right)^{x+3}$$

This limit involves an exponential function where both the base and the exponent are functions of x. As x approaches infinity, the base approaches 1, and the exponent also approaches infinity. This form is an indeterminate form of the type 1^∞, which requires careful manipulation to solve.

Initial Transformation

To solve this limit, we aim to rewrite the expression in a form that we can easily evaluate. A common technique is to express the function in terms of the exponential function e because we know the standard limit:

$$\lim_{x\to \infty} \left(1 + \frac{k}{x}\right)^x = e^k$$

So, let's start by manipulating the given expression to fit this form.

Step 1: Rewriting the Base

The key idea is to rewrite the fraction (x-4)/(x+1) in the form 1 + (something). We can do this by adding and subtracting 1 in the numerator or by performing long division. Here’s the process:

$$\frac{x-4}{x+1} = \frac{(x+1) - 5}{x+1} = 1 + \frac{-5}{x+1}$$

This manipulation allows us to express the original function as:

$$\lim_{x\to \infty} \left( 1 + \frac{-5}{x+1} \right)^{x+3}$$

Step 2: Adjusting the Exponent

Now, we want to make the exponent look like the reciprocal of the fraction inside the parentheses. We have (x+3) in the exponent, and we want it to be similar to (x+1). We can rewrite the exponent as follows:

$$x+3 = (x+1) + 2$$

So, our limit now looks like:

$$\lim_{x\to \infty} \left( 1 + \frac{-5}{x+1} \right)^{(x+1)+2}$$

Step 3: Separating the Exponential Term

Using the properties of exponents, we can separate the exponential term into two parts:

$$\lim_{x\to \infty} \left( 1 + \frac{-5}{x+1} \right)^{x+1} \cdot \left( 1 + \frac{-5}{x+1} \right)^{2}$$

This separation allows us to focus on the part that resembles the standard limit involving e.

Evaluating the Limit

Now that we have the expression in a suitable form, we can evaluate the limit by considering each part separately.

Step 4: Evaluating the First Part

The first part of the limit is:

$$\lim_{x\to \infty} \left( 1 + \frac{-5}{x+1} \right)^{x+1}$$

Let's make a substitution to clarify this limit further. Let u = x + 1. As x approaches infinity, u also approaches infinity. So, we can rewrite the limit as:

$$\lim_{u\to \infty} \left( 1 + \frac{-5}{u} \right)^{u}$$

This is now in the standard form for the limit definition of e, specifically:

$$\lim_{u\to \infty} \left( 1 + \frac{k}{u} \right)^{u} = e^k$$

In our case, k = -5, so the limit is:

$$\lim_{u\to \infty} \left( 1 + \frac{-5}{u} \right)^{u} = e^{-5}$$

Step 5: Evaluating the Second Part

The second part of the limit is:

$$\lim_{x\to \infty} \left( 1 + \frac{-5}{x+1} \right)^{2}$$

As x approaches infinity, the fraction -5/(x+1) approaches 0. Therefore, we have:

$$\lim_{x\to \infty} \left( 1 + \frac{-5}{x+1} \right)^{2} = (1 + 0)^2 = 1^2 = 1$$

Step 6: Combining the Results

Now we combine the results from evaluating both parts of the limit:

$$\lim_{x\to \infty} \left( 1 + \frac{-5}{x+1} \right)^{x+1} \cdot \left( 1 + \frac{-5}{x+1} \right)^{2} = e^{-5} \cdot 1 = e^{-5}$$

Final Answer

Therefore, the limit of the given function as x approaches infinity is:

$$\lim_{x\to \infty} \left( \frac{x-4}{x+1} \right)^{x+3} = e^{-5}$$

Alternative Approach: Using Natural Logarithms

Another powerful method to solve limits of this type is by using natural logarithms. This approach can simplify the expression and make the limit evaluation more straightforward.

Step 1: Introduce Natural Logarithms

Let's denote the given limit as L:

$$L = \lim_{x\to \infty} \left( \frac{x-4}{x+1} \right)^{x+3}$$

Take the natural logarithm of both sides:

$$\ln(L) = \ln \left[ \lim_{x\to \infty} \left( \frac{x-4}{x+1} \right)^{x+3} \right]$$

Since the natural logarithm is a continuous function, we can move the limit inside the logarithm:

$$\ln(L) = \lim_{x\to \infty} \ln \left[ \left( \frac{x-4}{x+1} \right)^{x+3} \right]$$

Using the logarithm power rule, we can bring down the exponent:

$$\ln(L) = \lim_{x\to \infty} (x+3) \ln \left( \frac{x-4}{x+1} \right)$$

Step 2: Rewrite the Expression

Now, we have a limit of the form ∞ * 0, which is another indeterminate form. To resolve this, we rewrite the expression as a fraction:

$$\ln(L) = \lim_{x\to \infty} \frac{\ln \left( \frac{x-4}{x+1} \right)}{\frac{1}{x+3}}$$

This transforms the limit into the form 0/0, which is suitable for applying 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 0/0 or ∞/∞, we can take the derivatives of the numerator and the denominator separately and then evaluate the limit:

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

In our case, f(x) = ln((x-4)/(x+1)) and g(x) = 1/(x+3). Let's find their derivatives.

First, find the derivative of f(x):

$$f'(x) = \frac{d}{dx} \left[ \ln \left( \frac{x-4}{x+1} \right) \right]$$

Using the chain rule:

$$f'(x) = \frac{1}{\frac{x-4}{x+1}} \cdot \frac{d}{dx} \left( \frac{x-4}{x+1} \right)$$

Now, find the derivative of (x-4)/(x+1) using the quotient rule:

$$\frac{d}{dx} \left( \frac{x-4}{x+1} \right) = \frac{(x+1)(1) - (x-4)(1)}{(x+1)^2} = \frac{x+1-x+4}{(x+1)^2} = \frac{5}{(x+1)^2}$$

Substitute this back into f'(x):

$$f'(x) = \frac{x+1}{x-4} \cdot \frac{5}{(x+1)^2} = \frac{5}{(x-4)(x+1)}$$

Next, find the derivative of g(x):

$$g'(x) = \frac{d}{dx} \left[ \frac{1}{x+3} \right] = -\frac{1}{(x+3)^2}$$

Now, apply L'Hôpital's Rule:

$$\ln(L) = \lim_{x\to \infty} \frac{\frac{5}{(x-4)(x+1)}}{-\frac{1}{(x+3)^2}} = \lim_{x\to \infty} \frac{5(x+3)^2}{-(x-4)(x+1)}$$

Step 4: Simplify the Limit

Expand and simplify the expression:

$$\ln(L) = \lim_{x\to \infty} \frac{5(x^2 + 6x + 9)}{-(x^2 - 3x - 4)}$$

Divide both the numerator and the denominator by x^2:

$$\ln(L) = \lim_{x\to \infty} \frac{5(1 + \frac{6}{x} + \frac{9}{x^2})}{-(1 - \frac{3}{x} - \frac{4}{x^2})}$$

As x approaches infinity, the terms with x in the denominator approach 0:

$$\ln(L) = \frac{5(1 + 0 + 0)}{-(1 - 0 - 0)} = -5$$

Step 5: Solve for L

Now we have ln(L) = -5. To find L, take the exponential of both sides:

$$L = e^{-5}$$

Conclusion

In this comprehensive guide, we explored two distinct methods to calculate the limit of ((x-4)/(x+1))^(x+3) as x approaches infinity. Both the initial transformation method and the natural logarithm method, coupled with L'Hôpital's Rule, led us to the same result: e^(-5). Understanding these methods enhances your problem-solving skills in calculus and provides valuable insights into handling indeterminate forms.

Whether you're a student tackling calculus problems or a math enthusiast eager to expand your knowledge, mastering these techniques will undoubtedly prove beneficial. Remember, practice is key, so try applying these methods to similar problems to solidify your understanding.

For further learning and exploration of limits and exponential functions, you might find valuable resources on websites like Khan Academy's Calculus Section.