Question

\(\operatorname{Lt}_{x \rightarrow-3} \frac{x^{2}-2 x-15}{x^{2}+2 x-3}=\) (1) 1 (2) 2 (3) \(-3\) (4) \(-4\)

Short answer

Answer: (2) 2

Step-by-step solution

Factor the numerator and denominator

We first factor the numerator and denominator of the given expression: Numerator: \(x^2-2x-15 = (x-5)(x+3)\) Denominator: \(x^2+2x-3 = (x+3)(x-1)\) So, we can write the expression as: \(\frac{(x-5)(x+3)}{(x+3)(x-1)}\)

Simplify the expression

Notice that \((x+3)\) is a common factor in both the numerator and the denominator. We can cancel this common factor: \(\frac{(x-5)(x+3)}{(x+3)(x-1)} = \frac{x-5}{x-1}\)

Calculate the limit

Now we calculate the limit of the simplified expression as x approaches -3: \(\operatorname{Lt}_{x \rightarrow -3} \frac{x-5}{x-1}\) Plug in x = -3: \(\frac{-3-5}{-3-1} = \frac{-8}{-4} = 2\) The limit as x approaches -3 is 2. The correct answer is (2) 2.

Key concepts

Algebraic Expressions

Algebraic expressions are the building blocks of algebra. They are made up of constants, variables, and arithmetic operations like addition, subtraction, multiplication, and division. In the given exercise, our main focus is the algebraic expression \( \frac{x^2-2x-15}{x^2+2x-3} \). This expression is a rational function, with a polynomial in both the numerator and the denominator.
Understanding the role of each term in an algebraic expression is essential. The term \(x^2\) represents the squared variable, indicating the parabola-like behavior of the graph at higher values. The linear terms \(-2x\) and \(2x\) help shape the curve, and the constants \(-15\) and \(-3\) shift the graph up or down. Recognizing these components can help you easily manipulate and simplify complex expressions.

Factoring Polynomials

Factoring polynomials is a key technique in simplifying algebraic expressions and solving equations. It involves rewriting a polynomial as a product of its factors, which are simpler expressions.
For example, in the exercise, we need to factor both the numerator and the denominator:

• Numerator: \(x^2-2x-15\) becomes \((x-5)(x+3)\)
• Denominator: \(x^2+2x-3\) becomes \((x+3)(x-1)\)

Identifying the common factors, such as \((x+3)\) in both the numerator and denominator, allows us to further simplify the given expression. We're essentially breaking down the polynomial into easier-to-handle pieces, which can later be canceled to simplify our calculations.

Indeterminate Forms

Indeterminate forms often occur when evaluating limits in calculus. At first glance, direct substitution of \(-3\) into the original expression \( \frac{x^2-2x-15}{x^2+2x-3} \) gives the form \( \frac{0}{0} \), which is undefined or an indeterminate form. To resolve this, further manipulation and simplification are needed.
This is where factoring plays a crucial role. By factoring and canceling the common terms, as shown in the solution where \((x+3)\) was canceled, we transform the indeterminate form into a determinate one, making it possible to compute the limit accurately. Handling indeterminate forms requires careful manipulation to avoid errors in the calculation process.

Simplifying Fractions

Simplifying fractions involves reducing them to their simplest form so they become easier to handle. This means that you look for common factors in the numerator and the denominator, which can be canceled out.
In the exercise, after factoring the polynomial expressions in the numerator and the denominator, we obtain \( \frac{(x-5)(x+3)}{(x+3)(x-1)} \). Here, the term \((x+3)\) appears in both the numerator and the denominator. By canceling these common factors, the fraction simplifies to \( \frac{x-5}{x-1} \).
Once we have the simplified fraction, calculating limits becomes straightforward. This process is crucial in calculus because it helps to convert complex expressions into manageable forms for evaluation.