Evaluating Limits
Evaluate lim(x to 2) of (3x + 1)
7
What technique do you use when direct substitution gives 0/0 and there is a square root in the expression?
rationalizing, multiply by the conjugate
What are the three conditions required for a function to be continuous at a point?
f(c) is defined, the limit exists at c, and the limit equals f(c)
Write the limit definition of the derivative.
f'(x) = lim(h to 0) of [f(x+h) - f(x)] / h
What does it mean for a limit to not exist (DNE)?
the left and right limits disagree, or the function grows without bound
If f(x) = x + 2 for x less than 1, and f(x) = 5 for x at least 1, find lim(x to 1 from the left) f(x)
3
Evaluate lim(x to 3) of (x^2 - 9)/(x - 3).
6, factor and cancel
A hole in a graph where the limit exists but does not equal the function value is what kind of discontinuity?
removable
Using the definition, find f'(x) for f(x) = x^2.
2x
True or false: every function that is differentiable at a point is also continuous there.
true
If lim(x to 3 from the left) f(x) = 4 and lim(x to 3 from the right) f(x) = 4, does lim(x to 3) f(x) exist, and what is it?
yes, it exists and equals 4
valuate lim(x to 4) of (sqrt(x) - 2)/(x - 4).
1/4, rationalize
A discontinuity where the left and right limits are different finite values is what kind?
jump, a nonremovable discontinuity
Differentiate f(x) = x^5
5x^4
Find f'(x) for f(x) = x^2 - 4x using the limit definition.
2x - 4
Evaluate lim(x to 0 from the left) of (1/x).
-infinite
Evaluate lim(x to 2) of [1/x - 1/2] / (x - 2).
-1/4, simplify the complex fraction
For f(x) = (x^2 - 1)/(x - 1), is this continuous at x = 1? If not, classify the discontinuity and state the value that would fill the hole.
not continuous; removable; the hole value is 2
Differentiate f(x) = 1/x^2 (rewrite with a negative exponent first).
-2/x^3
A function f has lim(x to 5) f(x) = 10, but f(5) = 7. Is f continuous at x = 5? What kind of discontinuity is this, and is it removable?
no; it is a removable discontinuity, the hole value is 10, redefining f(5) = 10 fixes it
f(x) = x^2 - 1 for x less than 2, and f(x) = 3x - 3 for x at least 2. Find lim(x to 2 from the left), lim(x to 2 from the right), and state whether lim(x to 2) f(x) exists.
both one-sided limits equal 3, so lim(x to 2) f(x) = 3
Evaluate lim(x to 1) of (x^3 - 1)/(x^2 - 1).
3/2, factor both as (x-1)(x^2+x+1) over (x-1)(x+1), cancel, then substitute x = 1
For what value of k is f(x) = x^2 + k for x less than 3, and f(x) = 5x - 4 for x at least 3, continuous at x = 3?
k = 2
Find the derivative of f(x) = 2x^3 - 5x^2 + x - 7 using the Power Rule, then evaluate f'(-1).
f'(x) = 6x^2 - 10x + 1; f'(-1) = 17
A function has f(2) = 5 and f'(2) = -3. Write the equation of the tangent line to f at x = 2.
y = -3x + 11