Problem 22 - Calculus Early Transcendentals

Question

Use the precise definition of a limit to prove the following limits. $$\lim _{x \rightarrow 3} \frac{x^{2}-7 x+12}{x-3}=-1$$

Step-by-Step Solution

Answer

Short Answer: To prove the given limit using the precise definition of a limit, we first simplified the given function and the limit value, resulting in the expression: $$f(x) - (-1) = x-3$$ Next, we chose a suitable value for delta, which turned out to be equal to epsilon. After verifying that our choice of delta satisfies the precise definition of a limit, we concluded that the given limit does indeed exist and is equal to -1.

Step 1: Understand the function and the limit value

$$\lim_{x \rightarrow 3} \frac{x^2-7x+12}{x-3}=-1$$ We are given the function $$f(x)=\frac{x^2-7x+12}{x-3}$$ and trying to prove the limit exists as x approaches 3.

Step 2: Understand the precise definition of a limit

We need to prove that for every \(\epsilon>0\) there exists a \(\delta > 0\) such that if \(0<\vert x-3\vert <\delta\) then \(\vert f(x) - (-1)\vert <\epsilon\).

Step 3: Find a relationship between f(x) and the limit value

Notice that $$f(x) - (-1) = \frac{x^2-7x+12}{x-3} + 1$$ Now, we can simplify this expression by finding a common denominator. $$\frac{x^2-7x+12+1(x-3)}{x-3}=\frac{x^2-6x+9}{x-3}=\frac{(x-3)^2}{x-3}$$ So, $$f(x) - (-1) = x-3$$

Step 4: Find a suitable value for \(\delta\)

Now we want to find a suitable value for \(\delta\) such that if \(0<\vert x-3\vert <\delta\), then \(\vert f(x) - (-1)\vert <\epsilon\). From the previous step, we found that \(f(x) - (-1) = x-3\). So, we have: $$\vert x - 3 \vert < \epsilon$$ Since this is already in the desired form, we can choose \(\delta\) to be equal to \(\epsilon\). Now, let's verify that our choice of \(\delta\) satisfies the precise definition of a limit.

Step 5: Verify the precise definition of a limit with the chosen value of \(\delta\)

We need to verify that for every \(\epsilon>0\) there exists a \(\delta > 0\) such that if \(0<\vert x-3\vert <\delta\) then \(\vert f(x) - (-1)\vert <\epsilon\). We have chosen \(\delta = \epsilon\). Let's prove it: 1. Let \(\epsilon > 0\) 2. Let \(\delta = \epsilon\) 3. Let \(0<\vert x-3\vert <\delta\) 4. Then \(\vert f(x) - (-1)\vert = \vert x - 3\vert <\delta=\epsilon\) All the conditions in the precise definition of a limit are satisfied. Therefore, we have proved using the precise definition of a limit that $$\lim_{x \rightarrow 3} \frac{x^2-7x+12}{x-3}=-1$$

Key Concepts

Epsilon-Delta Definition

The Epsilon-Delta definition is at the heart of understanding limits in calculus. This definition provides a formal way to prove the existence of a limit. It says that a function \( f(x) \) approaches a limit \( L \) as \( x \) approaches \( a \) if:

• For every real number \( \epsilon > 0 \), there exists a real number \( \delta > 0 \).
• Whenever \( 0 < |x-a| < \delta \), it follows that \( |f(x) - L| < \epsilon \).

This might sound a bit complex at first, but it simply means that as \( x \) gets closer and closer to \( a \), \( f(x) \) should get closer and closer to \( L \). In practice, to prove a limit using this definition, you'll spend a lot of time thinking about the relationship between \( x-a \) and \( f(x)-L \). This approach helps ensure that no matter how small we want \( f(x)-L \) to be, we can make it happen by choosing \( \delta \) appropriately.

Rational Functions

Rational functions are mathematical expressions that consist of a ratio of two polynomials. They look like \( \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomials. These functions have interesting characteristics:

• If the polynomial in the denominator, \( Q(x) \), is zero at any point, the rational function becomes undefined at that point.
• Rational functions can have holes, which occur when both the numerator and the denominator share a common factor.
• They can also have vertical asymptotes at points where \( Q(x) = 0 \) and the numerator is not zero.

In our exercise, the function \( \frac{x^2-7x+12}{x-3} \) represents a rational function. Here, the polynomial in the denominator, \( x-3 \), is zero when \( x = 3 \). But interestingly, when you simplify \( \frac{x^2-7x+12}{x-3} \) by factoring the numerator, you'll notice a removable discontinuity where the function reduces to \( x-3 \), revealing a simplification process crucial for proving limits.

Limit Proofs

Limit proofs, particularly using the Epsilon-Delta definition, can sometimes be intimidating, but with practice, they become manageable. The key steps involved are:

• Simplifying the expression like in our example \( f(x) - (-1) = x-3 \), which helps relate the function closely to a simple linear term.
• Choosing a value for \( \delta \) that typically mirrors or equates to \( \epsilon \) based on your simplification.
• Demonstrating that for every \( \epsilon > 0 \), you can find \( \delta > 0 \) such that the conditions of the Epsilon-Delta definition are satisfied.

In the given proof, after simplification, the relationship \( \vert f(x) - (-1)\vert = |x-3| \) naturally leads to choosing \( \delta = \epsilon \). This illustrates an elegant aspect of limit proofs - they often direct you towards the right choice for \( \delta \) through careful algebraic manipulation. Applying these steps accurately confirms the limit \( \lim_{x \rightarrow 3} \frac{x^2-7x+12}{x-3}=-1 \) truly through thoughtful examination.