UNIT 1 EXAM - LIMITS and CONTINUITY

Practice flashcards covering limits, continuity, piecewise functions, Intermediate Value Theorem, Squeeze Theorem, and graphical analysis from Math 1A Fall 2024 Unit 1 Exam.

The limit of (x+h)²+3(x+h)+4−(x²+3x+4) / h as h approaches 0 is __.

2x+3

From the given graph of function g, lim g(x) as x approaches 2 from the left is __.

1

Based on the graph of function g, the function g is __ at x=2.

not continuous

The limit of (sin 7x) / (sin 9x) as x approaches 0 is __.

7/9

From the given graph of function g, lim g(x) as x approaches 1 __.

does not exist (because the left-hand limit is -1 and the right-hand limit is 1)

If a rational function has a factor (x-c) in both the numerator and denominator, it indicates a __ at x=c.

hole

If a rational function has a factor (x-a) only in the denominator (and x=a is not a hole), it indicates a __ at x=a.

vertical asymptote

If y = (x-3)/(x-2) after simplification, the original function y = (x+b)(x+c) / ((x-3)(x-2)) implies that b + c = __.

-5

For the piecewise function f(x) = {3x+2, x

5

For a function f(x) to be continuous at x=c, three conditions must be met: f(c) must exist, lim f(x) as x approaches c must exist, and __.

lim f(x) as x approaches c must equal f(c)

For the function g(x) from the graph, it is discontinuous at x=2 because while g(2)=2, lim g(x) as x approaches 2 is 4, meaning __.

lim g(x) as x approaches 2 does not equal g(2)

Given the graph of g(x) and the piecewise function h(x) = {x^2+2x-3, x

-2/3

If a function f(x) is continuous on [a, b] and f(a) ≠ f(b), then for any value N between f(a) and f(b), there exists at least one c in (a, b) such that __.

f(c)=N

Given f is continuous, f(-6)=5, f(-2)=-1, and f(x)=0 for exactly one x, this x must be between __.

-6 and -2

To make the piecewise function f(x) = {x^2-2x+3, x

3

If f(x) = g(x) / h(x), where g and h are continuous for all real numbers, and lim f(x) = infinity as x approaches c, then x=c is a __.

vertical asymptote

If g(x)

L

Given g(x) = sin((pi/2)x) + 4 and h(x) = (-1/4)x^3 + (3/2)x + 9/4 satisfy g(x)

5

For the piecewise function f(x) = {x-3, x>=5; 12/(x-2), x

2

For the piecewise function f(x) = {x-3, x>=5; 12/(x-2), x<5}, the function is discontinuous at x=5 because the left-hand limit __________.

does not equal f(5)

From the given graph, lim f(x) as x approaches 4 from the left is __.

2

From the given graph, lim f(x) as x approaches 4 is __.

2

For the function shown in the graph, it is __ at x=4 due to a hole in the graph.

not continuous

To evaluate lim (sqrt(x+10)-3) / (x+1) as x approaches -1, one strategy is to multiply the numerator and denominator by the __ of the numerator, which is sqrt(x+10)+3.

conjugate

The limit lim (sqrt(x+10)-3) / (x+1) as x approaches -1 is __.

1/6

Given lim f(x) = -3 and lim g(x) = 9 as x approaches a, the limit lim (2f(x)) / (g(x)-f(x)) as x approaches a is __.

-1/2

From the given graph of g(x), lim g(x) as x approaches 2 __.

does not exist (as the left-hand limit is 1 and the right-hand limit is 3)

Given f(x) from graph and g(x-5). To find lim[f(x)+g(x-5)] as x->1, the left-hand limit evaluates to -1 and the right-hand limit evaluates to -1, which means the main limit is __.

-1

If functions j(x) and k(x) are continuous for all real numbers, then their composition j(k(x)) is also __.

continuous

Given h(x)=j(k(x))+5, with j and k continuous, h(1)=7, and h(7)=-5. To show there must be a value s for 1

1

To evaluate lim (x^3-8)/(x-2) as x approaches 2, we can factor the numerator using the difference of cubes formula a^3-b^3 = (a-b)(a^2+ab+b^2). The factored form of x^3-8 is __.

(x-2)(x^2+2x+4)

After factoring and canceling, lim (x^3-8)/(x-2) as x approaches 2 simplifies to evaluating lim (x^2+2x+4) as x approaches 2, which yields __.

12