Problem 7 - Technical Mathematics with Calculus

Question

Evaluate each limit. $$\lim _{x \rightarrow 1} \frac{x^{2}+2 x-3}{x-1}$$

Step-by-Step Solution

Answer

The limit is 4.

Step 1: Factor the numerator

Begin by factoring the quadratic expression in the numerator. The quadratic expression can be factored as \(x^2+2x-3=(x+3)(x-1)\).

Step 2: Cancel common factors

Cancel the \(x-1\) terms from the numerator and the denominator because they are common factors and the limit exists for \(x eq 1\): \[\lim _{x \rightarrow 1} \frac{x^{2}+2 x-3}{x-1} =\lim _{x \rightarrow 1} \frac{(x+3)(x-1)}{x-1}\]. After cancellation, we are left with \[\lim _{x \rightarrow 1} (x+3)\].

Step 3: Evaluate the limit

Now evaluate the simplified expression directly by substituting \(x = 1\): \[\lim _{x \rightarrow 1} (x+3) = (1+3) = 4\].

Key Concepts

Factoring Polynomials

Understanding how to factor polynomials is crucial in simplifying complex algebraic expressions, especially when working with limits. Polynomials are expressions that consist of variables and coefficients, involving only the operations of addition, subtraction, multiplication, and non-negative integer exponents. In the exercise, the numerator of the given limit is a quadratic polynomial. Factoring this polynomial means finding two binomials that multiply together to give the original quadratic.

To factor a quadratic polynomial like \(x^2 + 2x - 3\), we look for two numbers that add up to the coefficient of the middle term, which is 2, and multiply to the constant term, which is -3. In our case, these numbers are 3 and -1. This gives us the factorization \((x + 3)(x - 1)\).

Mastering the skill of factoring not only simplifies calculations but also enables us to see patterns and relationships within algebraic expressions, allowing for easier manipulation and solution finding.

Canceling Common Factors

When simplifying rational expressions, canceling common factors is a technique often utilized to reduce the expression to its lowest terms. This method is particularly important when evaluating limits. A rational expression is made up of a numerator and denominator that are both polynomials.

In the solution provided, the quadratic polynomial from the numerator, after factoring, shares a common factor with the denominator: \((x - 1)\). Since we want to evaluate the limit as \(x\) approaches 1, and not the value of the function at \(x = 1\) itself, we can safely cancel out the common factors, provided \(x\) is not equal to 1. This is because as \(x\) moves closer to 1, the factors other than \(x - 1\) determine the behavior of the expression.

It’s akin to removing redundant parts of an equation to see what’s truly affecting the output as we hone in on the point of interest, in this case, \(x = 1\). This process simplifies the expression and makes evaluation of the limit feasible.

Direct Substitution in Limits

Direct substitution is a straightforward way to evaluate limits when a function is continuous at the point to which \(x\) is approaching. Once we’ve simplified the original expression by factoring and canceling common factors, as in the steps described above, we reach a point where the function's continuity allows for a direct substitution.

In our example, after simplification, we obtained \(\frac{{x^{2}+2 x-3}}{{x-1}} = x+3\). Because there are no more common factors and the function is continuous at \(x = 1\), we can replace \(x\) with 1 to find the limit. This makes the process of finding the limit as \(x\) approaches 1 incredibly efficient: \(\text{lim}_{x \rightarrow 1} (x+3) = (1+3) = 4\).

This step concludes that as \(x\) approaches 1, the original function also approaches the value 4. It's important to remember that direct substitution is only valid when there are no discontinuities, such as holes or asymptotes, at that point in the function.