From AP Live Calculus AB – April 13; Unit 1: Limits And Continuity 1-3: Use The Graph Of The Piecewise Function, F, Shown To Find The Requested Values. 1. F(0) En 2. Lim F(X) (3. 1 3. Lim F(X) X – 3 4. X + 5 X=-5 X2 – 25 Lim – 2x +3 – 3 5. Lim X+ 3x – 3
The Correct Answer and Explanation is :
To evaluate the given limits and function values, we need to analyze the piecewise function ( f(x) ) based on its graph. Here’s how to approach each part:
- ( f(0) ): – ocate ( x = 0 ) on the graph. – dentify the value of ( f(x) ) at this point. – f the graph shows a filled circle at ( x = 0 ), then ( f(0) ) is the value at that point. – f there’s an open circle, ( f(0) ) is undefined.
- ( \lim_{x \to 3} f(x) ): – xamine the behavior of ( f(x) ) as ( x ) approaches 3 from both the left and right. – f the function approaches the same value from both sides, that value is the limit. – f the function approaches different values from each side, the limit does not exist.
- ( \lim_{x \to 3^-} f(x) ): – ocus on the left-hand behavior of ( f(x) ) as ( x ) approaches 3. – dentify the value the function approaches from the left side.
- ( \lim_{x \to -5} \frac{x + 5}{x^2 – 25} ): – actor the denominator: ( x^2 – 25 = (x + 5)(x – 5) ). – implify the expression: ( \frac{x + 5}{(x + 5)(x – 5)} = \frac{1}{x – 5} ). – s ( x ) approaches -5, the expression becomes ( \frac{1}{-5 – 5} = \frac{1}{-10} ). – herefore, ( \lim_{x \to -5} \frac{x + 5}{x^2 – 25} = -\frac{1}{10} ).
- ( \lim_{x \to 3} \frac{\sqrt{2x + 3} – 3}{x – 3} ): – irect substitution leads to the indeterminate form ( \frac{0}{0} ). – ultiply the numerator and denominator by the conjugate of the numerator: [ \frac{\sqrt{2x + 3} – 3}{x – 3} \times \frac{\sqrt{2x + 3} + 3}{\sqrt{2x + 3} + 3} = \frac{(2x + 3) – 9}{(x – 3)(\sqrt{2x + 3} + 3)} = \frac{2x – 6}{(x – 3)(\sqrt{2x + 3} + 3)} ] – actor the numerator: [ \frac{2(x – 3)}{(x – 3)(\sqrt{2x + 3} + 3)} ] – ancel the ( (x – 3) ) terms: [ \frac{2}{\sqrt{2x + 3} + 3} ] – ubstitute ( x = 3 ): [ \frac{2}{\sqrt{2(3) + 3} + 3} = \frac{2}{\sqrt{9} + 3} = \frac{2}{3 + 3} = \frac{2}{6} = \frac{1}{3} ] – herefore, ( \lim_{x \to 3} \frac{\sqrt{2x + 3} – 3}{x – 3} = \frac{1}{3} ).
For a more in-depth understanding of limits and continuity, you might find the following video helpful:
video2022 Live Review 3 | AP Calculus AB | How to Find Limits & Continuityturn0search2