Question

business calculus review
business calculus review worksheet

1. evaluate the limit: \\(\lim\limits_{x\to 4}(3x^2 - 2x + 5)\\)

1. evaluate: \\(\lim\limits_{x\to 3}\frac{x^2 - 9}{x - 3}\\)

Explanation:

Problem 1:

Step1: Substitute \( x = 4 \) into the function

For the limit \( \lim_{x \to 4}(3x^2 - 2x + 5) \), since the function \( 3x^2 - 2x + 5 \) is a polynomial, we can use the direct substitution property of limits (polynomial functions are continuous everywhere, so the limit as \( x \) approaches a value is the function's value at that point). Substitute \( x = 4 \) into the function:
\( 3(4)^2 - 2(4) + 5 \)

Step2: Calculate the value

First, calculate \( 4^2 = 16 \). Then:
\( 3\times16 - 2\times4 + 5 = 48 - 8 + 5 \)
\( 48 - 8 = 40 \), and \( 40 + 5 = 45 \)

Step1: Factor the numerator

The numerator \( x^2 - 9 \) is a difference of squares, which factors as \( (x - 3)(x + 3) \). So the limit becomes:
\( \lim_{x \to 3}\frac{(x - 3)(x + 3)}{x - 3} \)

Step2: Cancel the common factor

For \( x
eq 3 \) (since we are taking the limit as \( x \) approaches 3, not evaluating at \( x = 3 \)), we can cancel the \( (x - 3) \) terms:
\( \lim_{x \to 3}(x + 3) \)

Step3: Substitute \( x = 3 \)

Now substitute \( x = 3 \) into \( x + 3 \):
\( 3 + 3 = 6 \)

Answer:

\( 45 \)

Problem 2: