Problem 42 - Calculus Early Transcendentals

Question

Evaluate the following limits, where a and \(b\) are fixed real numbers. \(\lim _{t \rightarrow 2} \frac{3 t^{2}-7 t+2}{2-t}\)

Step-by-Step Solution

Answer

Question: Evaluate the following limit, where a and b are fixed real numbers: \(\lim_{t\rightarrow 2}\frac{3t^2 - 7t + 2}{2-t}\) Answer: -5

Step 1: Observe the given expression and try to simplify it

We are given the expression: \(\lim _{t \rightarrow 2} \frac{3 t^{2}-7 t+2}{2-t}\) First, notice that we can't directly plug in \(t=2\) since it would yield a denominator of zero. So, we need to find a way to simplify the expression before evaluating the limit.

Step 2: Factor the numerator

Looking at the numerator, we can attempt to factor it: $$3 t^{2}-7 t+2 = (3t-1)(t-2)$$ So, the expression becomes: \(\lim _{t \rightarrow 2} \frac{(3t-1)(t-2)}{2-t}\)

Step 3: Factor out negative one in the denominator

Notice that we can factor out \((-1)\) from the denominator, so it becomes: $$\lim _{t \rightarrow 2} \frac{(3t-1)(t-2)}{-(t-2)}$$

Step 4: Simplify the expression

Now, we can cancel out the common term \((t-2)\) from the numerator and the denominator: $$\lim _{t \rightarrow 2} \frac{(3t-1)}{-1}$$

Step 5: Insert the limit value

After simplifying the expression, we can now plug in the limit value \(t=2\): $$\lim _{t \rightarrow 2} \frac{(3(2)-1)}{-1} =\frac{5}{-1}$$

Step 6: Evaluate the limit

Therefore, the limit is equal to \(\frac{-5}{1}\) or \(-5\): $$\lim _{t \rightarrow 2} \frac{3 t^{2}-7 t+2}{2-t} = -5$$

Key Concepts

Limits in Calculus

In calculus, a limit captures the behavior of a function as its argument approaches some value. It's a foundational concept that helps in studying the continuity, derivatives, and integrals of functions. When evaluating a limit, we're interested in finding the value that a function gets close to as the input gets close to some number, often without ever reaching that number exactly.

Take the example of evaluating \(\lim _{t \rightarrow 2} \frac{3 t^{2}-7 t+2}{2-t}\). The core challenge here is that direct substitution of \(t=2\) isn't possible as it leads to division by zero, a situation that indicates the need for further manipulation of the function to find its limit as \(t\) approaches 2.

Factoring Polynomials

Factoring polynomials is breaking down a polynomial into simpler 'factor' polynomials that, when multiplied together, give back the original polynomial. It's like reverse-engineering a multiplication problem. When factoring, we look for patterns such as the difference of squares, perfect square trinomials, and sum or difference of cubes, among others.

Factoring is particularly essential in limit evaluation if the polynomial in question causes an indeterminate form like \(0/0\). For the problem \(\lim _{t \rightarrow 2} \frac{3 t^{2}-7 t+2}{2-t}\), the numerator \(3 t^{2}-7 t+2\) was factored into \(3t-1)(t-2)\), which was a crucial step in simplifying the expression and solving for the limit.

Simplifying Expressions

Simplifying expressions in mathematics means to alter them into their most basic form without changing their value. This often involves reducing fractions, factoring, canceling out common factors, and applying algebraic rules. In our limit problem, after factoring the numerator and rewriting the denominator, simplifying means canceling the common factor of \(t-2\) from both the numerator and the denominator.

This process yields a simpler expression that can usually be evaluated through direct substitution or other means. The goal of simplification is to make complex or undefined expressions into something more manageable or clearly defined.

Direct Substitution in Limits

Direct substitution involves plugging the limit value directly into the function, provided it does not result in an undefined expression such as \(0/0\) or \(\infty/\infty\). In cases where the function simplifies to a point that it can accept the limit value without indeterminate forms, direct substitution is the easiest and fastest way to evaluate the limit.

In the example at hand, once we simplified the expression to \(\lim _{t \rightarrow 2} \frac{(3t-1)}{-1}\), we were able to directly substitute \(t=2\) to calculate the final limit value of \(\frac{-5}{1}\) or \(\-5\). This shows how substantial initial simplification can be in resolving what might initially appear to be a complex limit problem.