They indicate the behavior of the function as it approaches certain values.
lim (x → 0) (x^3 - 7)/(x^2 - 4); lim (x → ∞) (x^2 - 10)/(x^2 + 2)
It indicates the behavior of the function f(x) as x approaches 6.
If δ < |x - 1|, then |f(x) - 2| < 0.1.
a = 3, b = 2.
1) f(a) is defined; 2) The limit of f(x) as x approaches a exists; 3) The limit of f(x) as x approaches a equals f(a).
∞.
Use the product rule: f'(x) = u'v + uv', where u = (1 + 2x^2) and v = (3 - 5x^2).
It indicates the behavior of the function f(x) as x approaches infinity.
You MUST show your work.
Calculate f(x + h) = 4(x + h) + 3 and then use the limit definition.
The limit is undefined.
It helps to prove that a function has a root within a certain interval.
-7/5.
Use the chain rule: f'(x) = g'(h(x)) * h'(x).
It means that the function does not have a defined derivative at that point.
There are 2 values for which the function is not differentiable; the function is continuous from the left at x = 0.
It helps to find the limit of a function by 'squeezing' it between two other functions.
y = 4/3.
There exists at least one c in [1, 5] such that f(c) = 2.
By analyzing the limits and the defined points of the function.
f'(x) = lim (h -> 0) [(f(x + h) - f(x)) / h].
Apply the product rule: f'(x) = u'v + uv', where u = x^2 and v = e^(3x).
It indicates the behavior of the function f(x) as x approaches 1.
f(x) = 4x + 3.
f(x) = e^(-x) + x, f(x) = 7sin(x), f(x) = 6/(x^2 - 8), f(x) = 5/(x + 1)
Use the chain rule: f'(10) = g'(h(10)) * h'(10) = 0 * 35 = 0.
The function must not have any breaks or jumps in its domain.
f'(x) = 4.
g(x) is the derivative of f(x); g(x) is differentiable everywhere; g(x) is an odd function; f(x) is an even function.
