Question

Find the following limits or state that they do not exist. Assume \(a, b, c,\) and k are fixed real numbers. $$\lim _{t \rightarrow 2} \frac{3 t^{2}-7 t+2}{2-t}$$

Short answer

Answer: The limit of the expression as \(t\) approaches 2 is -5.

Step-by-step solution

Identify the expression's limit

We are asked to find the limit of the given expression: $$\lim _{t \rightarrow 2} \frac{3 t^{2}-7 t+2}{2-t}$$

Substitute the value of t if possible

Let's first see if we can directly substitute \(t=2\) in the expression. We will find: $$\lim _{t \rightarrow 2} \frac{3 (2)^{2}-7 (2)+2}{2-2} = \frac{12-14+2}{0}$$ This involves division by zero, so we cannot directly substitute the value of \(t\) into the expression.

Factor the numerator

Now, we must attempt to factor the numerator, to see if there is a common factor in the numerator and the denominator. Observe that \(3t^2 - 7t + 2\) may be factored as \((3t-1)(t-2)\). So, the expression becomes: $$\lim _{t \rightarrow 2} \frac{(3t-1)(t-2)}{2-t}$$

Factor out a -1 from the denominator

We will now remove the negative sign from the denominator, making the expression easier to manage. $$\lim _{t \rightarrow 2} \frac{(3t-1)(t-2)}{-1(t-2)}$$

Cancel out the common factors

Since both the numerator and the denominator have a common factor \((t-2)\), we can cancel it out from both of them. $$\lim _{t \rightarrow 2} \frac{3t-1}{-1}$$

Substitute the value of t again

Now, we can directly substitute the value of \(t=2\) in the simplified expression: $$\lim _{t \rightarrow 2} \frac{3(2)-1}{-1} = \frac{6-1}{-1} = \frac{5}{-1} = -5$$ So, the limit of the given expression as \(t\) approaches 2 is -5.

Key concepts

Direct Substitution

When working with limits, the first step is often trying to substitute the limit point directly into the function. This is known as direct substitution. The idea is simple: replace the variable with the value it's approaching and see if you get a real number. If you do, great! That's the limit.

However, if you end up with a situation like division by zero, as in our example, direct substitution alone doesn’t give us the answer. Here, substituting directly gave us a zero in the denominator, which is undefined in mathematics.

Therefore, when direct substitution fails due to circumstances like division by zero, we need to try other methods like factoring to resolve the issue.

Factoring

Factoring involves rewriting an expression as a product of simpler factors. This technique is especially useful when dealing with polynomials. In our example, the polynomial in the numerator was factored into two terms:

• \( (3t-1)(t-2) \)

Factoring can reveal common factors in both the numerator and the denominator that can be canceled out, simplifying the expression and resolving issues like division by zero.

It's often a go-to method when dealing with limits involving polynomials, especially when direct substitution doesn’t work. Always check factoring possibilities carefully to avoid missing potential simplifications.

Division by Zero

Encountering division by zero while evaluating limits often signals that direct substitution isn't going to work. This is because division by zero is undefined and makes the expression impossible to evaluate straightaway.

In our specific example, substituting \(t = 2\) led to the denominator being zero, which halted our direct substitution method. However, by factoring and manipulating the expression, we sometimes find that the zero divisor is part of a common factor that can be canceled out, allowing further evaluation.

Understanding this behavior is key because it shows us how limits that initially seem impossible to find due to division by zero can sometimes be simplified to a workable form.

Canceling Common Factors

Canceling common factors is a crucial technique in limit evaluation. This comes into play after the factoring step, where we identify and cancel the same term present in both the numerator and the denominator.

In this problem, once we factored the numerator, we noticed a common term \((t-2)\) in both the numerator and the denominator. By canceling this factor, we simplified the expression into something more manageable:

• \( \frac{3t-1}{-1} \)

This simplification resolved the division by zero issue and allowed us to substitute \(t=2\) directly, finding the limit easily.

Such techniques are helpful because they simplify calculations and prevent division errors, saving us from dead-end evaluations.