Asymptotes And Limits: A Detailed Solution

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Asymptotes and Limits: A Detailed Solution

Hey guys! Today, we're diving deep into a fantastic math problem that involves finding asymptotes and limits. These are fundamental concepts in calculus, and mastering them is crucial for understanding more advanced topics. We'll break down each step, so you can follow along and really grasp the ideas. So, let's get started!

(a) Finding Asymptotes of f(x) = (2x² + 1) / (4x² - 25)

When we talk about asymptotes, we're referring to lines that a function approaches but never quite touches. There are two main types we'll focus on here: vertical and horizontal asymptotes. Finding these lines helps us understand the function's behavior, especially at its extremes.

Vertical Asymptotes

Let's kick things off with vertical asymptotes. These occur where the function approaches infinity (or negative infinity). Practically, this usually happens where the denominator of a rational function equals zero. So, our first step is to figure out when the denominator, 4x² - 25, is zero. Guys, let's break it down:

  1. Set the denominator equal to zero: 4x² - 25 = 0
  2. Add 25 to both sides: 4x² = 25
  3. Divide by 4: x² = 25/4
  4. Take the square root of both sides: x = ±√(25/4)
  5. Simplify: x = ±5/2

So, we've found that the denominator is zero when x = 5/2 and x = -5/2. This tells us we potentially have vertical asymptotes at these points. To confirm, we need to make sure the numerator isn't also zero at these points (because if both numerator and denominator are zero, we might have a hole instead of an asymptote). Let's check the numerator, 2x² + 1:

  • When x = 5/2, the numerator is 2(5/2)² + 1 = 2(25/4) + 1 = 25/2 + 1 = 27/2, which is not zero.
  • When x = -5/2, the numerator is 2(-5/2)² + 1 = 2(25/4) + 1 = 25/2 + 1 = 27/2, which is also not zero.

Since the numerator is not zero at these points, we can confidently say that we have vertical asymptotes at x = 5/2 and x = -5/2. Awesome!

Horizontal Asymptotes

Next up, let's tackle the horizontal asymptotes. These tell us what happens to the function as x approaches positive or negative infinity. To find them, we'll look at the degrees of the polynomials in the numerator and the denominator.

Our function is f(x) = (2x² + 1) / (4x² - 25). Notice that the highest power of x in both the numerator and the denominator is x² (degree 2). When the degrees are the same, the horizontal asymptote is the ratio of the leading coefficients (the coefficients of the highest power terms).

In our case:

  • The leading coefficient in the numerator is 2.
  • The leading coefficient in the denominator is 4.

Therefore, the horizontal asymptote is y = 2/4, which simplifies to y = 1/2. So, as x gets incredibly large (positive or negative), the function approaches the line y = 1/2. Great job!

(b) Analyzing the Piecewise Function

Now, let's switch gears and look at the piecewise function:

f(x) = { (x² - 16) / (x² - 4x) , x ≠ 4; p , x = 4 }

This function behaves like a rational function everywhere except at x = 4, where its value is defined to be the constant p. The key here is to figure out what value of p would make the function continuous. To do this, we need to find the limit of the rational part as x approaches 4 and see if we can make p equal to that limit.

(i) Determining lim (x→4) f(x)

To find the limit as x approaches 4, we'll focus on the rational part of the function: (x² - 16) / (x² - 4x). If we try to directly substitute x = 4, we get (4² - 16) / (4² - 4*4) = 0/0, which is an indeterminate form. This means we need to do some algebraic manipulation to simplify the expression before we can evaluate the limit.

Guys, the trick here is to factor both the numerator and the denominator. Let's do it:

  • Numerator: x² - 16 is a difference of squares, so it factors as (x - 4)(x + 4).
  • Denominator: x² - 4x has a common factor of x, so it factors as x(x - 4).

Now, our function looks like this: f(x) = (x - 4)(x + 4) / x(x - 4). Notice that we have a common factor of (x - 4) in both the numerator and the denominator. Since we're considering the limit as x approaches 4 (but not actually equal to 4), we can cancel out this factor:

f(x) = (x + 4) / x, for x ≠ 4

Now, let's try substituting x = 4 into this simplified expression: (4 + 4) / 4 = 8 / 4 = 2. So, the limit as x approaches 4 of f(x) is 2. Fantastic!

Therefore, lim (x→4) f(x) = 2.

(ii) Discussion

Okay, so we've found that the limit of the function as x approaches 4 is 2. Now, let's discuss what this means in the context of the piecewise function and continuity.

Recall that our function is defined as:

f(x) = { (x² - 16) / (x² - 4x) , x ≠ 4; p , x = 4 }

We want to know if we can choose a value for p that makes this function continuous at x = 4. For a function to be continuous at a point, three things must be true:

  1. The function must be defined at that point (f(4) must exist).
  2. The limit of the function as x approaches that point must exist (lim (x→4) f(x) must exist).
  3. The value of the function at the point must equal the limit (f(4) = lim (x→4) f(x)).

Let's check these conditions for our function at x = 4:

  1. f(4) exists: The function is defined at x = 4, and f(4) = p.
  2. lim (x→4) f(x) exists: We found that lim (x→4) f(x) = 2.
  3. f(4) = lim (x→4) f(x): This is where the magic happens! For the function to be continuous, we need p to be equal to the limit, which is 2. So, we need p = 2.

In conclusion, the piecewise function f(x) is continuous at x = 4 if and only if p = 2. If p is any other value, the function will have a discontinuity (specifically, a removable discontinuity) at x = 4. Basically, by setting p = 2, we "fill in the hole" that would otherwise exist in the graph of the function at x = 4.

Wrapping Up

So, guys, we've tackled a pretty comprehensive problem involving asymptotes and limits! We found the vertical and horizontal asymptotes of a rational function, and we determined the limit of a piecewise function. We also discussed how to make a piecewise function continuous by carefully choosing the value of the constant. Hopefully, this breakdown has helped you solidify your understanding of these important calculus concepts. Keep practicing, and you'll become a pro in no time!