Problem 7 - Calculus: Early Transcendentals

Question

Let \(f(x)=\frac{x^{2}-4}{x-2}\) a. Calculate \(f(x)\) for each value of \(x\) in the following table. b. Make a conjecture about the value of \(\lim _{x \rightarrow 2} \frac{x^{2}-4}{x-2}\) $$\begin{array}{|l|l|l|l|l|}\hline x & 1.9 & 1.99 & 1.999 & 1.9999 \\\\\hline f(x)=\frac{x^{2}-4}{x-2} & & & & \\\\\hline x & 2.1 & 2.01 & 2.001 & 2.0001 \\\\\hline f(x)=\frac{x^{2}-4}{x-2} & & & & \\\\\hline\end{array}$$

Step-by-Step Solution

Answer

Answer: The conjectured value for the limit is 4.

Step 1: Compute f(x) for each x value

To calculate f(x) for each value in the table, plug in the x value into the function and simplify: $$f(x) = \frac{x^2 - 4}{x - 2}$$ For \(x = 1.9\): $$f(1.9) = \frac{(1.9)^2 - 4}{1.9 - 2} \approx 3.9$$ For \(x = 1.99\): $$f(1.99) = \frac{(1.99)^2 - 4}{1.99 - 2} \approx 3.99$$ For \(x = 1.999\): $$f(1.999) = \frac{(1.999)^2 - 4}{1.999 - 2} \approx 3.999$$ For \(x = 1.9999\): $$f(1.9999) = \frac{(1.9999)^2 - 4}{1.9999 - 2} \approx 3.9999$$ For \(x = 2.1\): $$f(2.1) = \frac{(2.1)^2 - 4}{2.1 - 2} \approx 4.1$$ For \(x = 2.01\): $$f(2.01) = \frac{(2.01)^2 - 4}{2.01 - 2} \approx 4.01$$ For \(x = 2.001\): $$f(2.001) = \frac{(2.001)^2 - 4}{2.001 - 2} \approx 4.001$$ For \(x = 2.0001\): $$f(2.0001) = \frac{(2.0001)^2 - 4}{2.0001 - 2} \approx 4.0001$$ Here is the completed table: $$\begin{array}{|l|l|l|l|l|}\hline x & 1.9 & 1.99 & 1.999 & 1.9999 \\\\\hline f(x)=\frac{x^{2}-4}{x-2} & 3.9 & 3.99 & 3.999 & 3.9999 \\\\\hline x & 2.1 & 2.01 & 2.001 & 2.0001 \\\\\hline f(x)=\frac{x^{2}-4}{x-2} & 4.1 & 4.01 & 4.001 & 4.0001 \\\\\hline\end{array}$$ #b. Make a conjecture about the value of \(\lim _{x \rightarrow 2}\frac{x^{2}-4}{x-2}\)

Step 2: Conjecture the limit based on the observations

Observing the values of \(f(x)\) in the table, we see that as \(x\) approaches 2, \(f(x)\) seems to be approaching 4. Therefore, we can conjecture that: $$\lim_{x\rightarrow2} \frac{x^2 - 4}{x - 2} = 4$$

Key Concepts

Limit of a Function

Understanding the limit of a function is crucial when delving into calculus, especially since it's the foundation for other advanced concepts like derivatives and integrals. A limit, in essence, describes the value a function approaches as the input variable, usually denoted as 'x,' gets close to a certain point.

Let's break this down with a tangible example. Imagine you're walking towards a door—this door represents a particular value 'a.' As you get closer to the door, you're not exactly there, but you can clearly see what you'll be facing once you open it; this is akin to the function's value getting close to the limit as 'x' approaches 'a.'

Mathematically, we write this as \( \lim_{x\rightarrow a} f(x) = L \), where 'L' signifies the limit. It's important to note that the limit does not always equate to the function's value at 'x = a.' In fact, there may be instances where 'f(a)' doesn't even exist, yet the limit as x approaches 'a' can still be a finite number or even infinity.

Rational Functions

Rational functions are akin to fractions, but instead of numbers, we have polynomials in the numerator and denominator. Expressed generically, a rational function is of the form \( f(x) = \frac{p(x)}{q(x)} \), where \( p(x) \) and \( q(x) \) are polynomials and \( q(x) \) is non-zero.

One of the intriguing aspects of rational functions is their behavior near points where the denominator goes to zero. These are known as singularities or poles. The function may indeed soar off to infinity, or strangely enough, it may settle into a finite limit. This often leads to scenarios that seem paradoxical initially but are perfectly logical within the framework of calculus.

A classic example is the function \( \frac{x^2-4}{x-2} \). At \( x = 2 \), we should see the denominator turning to zero. However, this particular function has a hole at \( x = 2 \) since \( x^2-4 \) is the difference of squares and factors as \( (x+2)(x-2) \). After simplification, we are left with \( x + 2 \) when \( x eq 2 \) which seems well-behaved and implies that near \( x = 2 \) the function should behave similarly to a straight line.

Approaching Values

The idea of values 'approaching' something is at the heart of the limit concept. When working out a limit problem, we're not typically interested in the function's value at a specific point, but rather what value the function is getting closer to as 'x' nears that point.

Approaching, however, should not be mistaken for reaching or equaling. We're content with getting as close as we wish to the point, mirroring how we can perpetually halve the distance to a wall without ever touching it—it's an endless pursuit getting infinitely closer, which is a core principle in understanding limits in calculus.

In the provided exercise, the function's values grow increasingly closer to 4 as 'x' moves towards 2, both from the left and right. It's like trying to predict the ending of a story by reading the pages just before the final chapter; we can make a reasonable guess based on what's approaching. In this case, regardless of whether 'x' is slightly less than or slightly more than 2, the 'ending' we predict for \( f(x) \) as \( x \) approaches 2 is 4. This intuitive method can be formalized through rigorous mathematical definitions and the use of precise ε-δ language in limit proofs.