Limits and Continuity

lim_x—af(x) = L if an only if…

lim_x—a^-f(x) = L = lim_x—a⁺f(x)

• solve each side, if not = - DNE

lim_x—af(x) = L can only exist if…

f(x) is defined on an open interval surrounding a (except possibly at a itself)

• continuous because

  • polynomial

  • logarithmic for positive #s

• write when IVT + identifying continuities

  • f(x) continuous, reason + reason

lim_x—ac =

c

evaluating a limit

• plug a into the function

  • finished if the result is a real number or +- infinity

    • limit DNE if +- infinity

  • limit DNE if the result is a real number divided by zero

  • limit indeterminate if the result is:

    • • 0/0

      • ∞/∞

      • (0)(∞)

      • ∞-∞

    • find another way to evaluate the limit

methods for evaluating indeterminate limits

• factor and cancel

• multiply by the conjugate (radicals + trig functions)

• simplify the complex fraction

• determining whether the num or denom approaches infinity faster as x — ∞

• Squeeze Theorem

• u-substitution

• divide by the largest degree as x — ∞ (rational functions)

determining whether the num or denom approaches infinity faster as x — ∞

• if the num approaches infinity faster than the denominator, the limit is ∞

• if the denom approaches infinity faster than the num, the limit is 0

• functions from approaching infinity the fastest to the lowest

  • exponential

  • polynomial

  • root

  • logarithmic

Squeeze Theorem

for functions defined on an open interval surrounding a, but not necessarily at a

if f(x) <= g(x) <= h(x) and lim_x—af(x) = L = lim_x—ah(x) = L, then lim_x—ag(x) = L

things to remember

• when multiplying an inequality by a negative, flip the inequality signs

  • when multiplying an inequality by something that is only negative for some portion of the domain of the function, split it up

• when evaluating lim_x—af(x) by looking at the graph of f(x), remember that the limit is the value that the function approaches, not necessarily the value of the function at a

lim_x—0sinx/x

lim_x—0x/sinx

1

lim_x—0tanx/x

lim_x—0x/tanx

1

lim_x—0(1-cosx)/x

lim_x—0(cosx-1)/x

0

vertical asymptotes

• if a function’s limit at a real number a is either positive or negative infinity, it has a vertical asymptote at a

• to determine how the function behaves on each side of a, simply plug in a number very close to a on both sides of a to determine whether it approaches positive or negative infinity on each side

  • small +, small -

horizontal asymptotes

• a horizontal asymptote to the right can be determined by finding lim_x—∞f(x)

• a horizontal asymptote to the left can be determined by finding lim_x—-∞f(x)

• you must check both positive and negative infinity when justifying horizontal asymptotes

• a rational function will always have the same horizontal asymptote to the left and to the right (but you must show both in a free response question)

  • divide each term by the largest degree

definition of continuity

for f(x) to be continuous at a point x = a

• f(x) must be defined at x = a

• lim_x—af(x) must exist

• f(a) = lim_x—af(x)

when using the definition of continuity to determine if a piecewise defined function is continuous at the endpoints of the separate pieces of its domain, make sure to use both one sided limits to determine if the limit at those values exists

intermediate value theorem

• if f is continuous on [a, b], then for every value w on (f(a), f(b)), there exists a value c on (a, b) such that f(c)= w

• the Intermediate Value Theorem is commonly used to find an interval on which f(x) has an x-intercept

• remember to state that the function is continuous on the given interval and make sure to also state the reason why it is continuous on that interval

  • write on problems:

    • f(x) = y

    • f(x2) = y2

    • f(x) =/ f(x2)

    • f(c) = w

    • y < w < y2

    • y < f(c) < y2

    • x < c < x

      • by intermediate value theorem, for every w on (y,y) there exists a c in (x,x) such that f(c) = w

    • by IVT f(x) has a solution on (x, x)

      • when proving a solution in a domain of a function

classifying discontinuities

• removable discontinuities occur x = c when c makes both the numerator and denominator of a function equal 0. Graphically, the function will have a hole at x = c in the graph. In addition to the hole at x = c, the function’s graph might still be defined at c but lim_x—cf(x) ≠ f(c)

• infinite discontinuities occur at x = c when the lim_x—c-f(x) and/or lim_x—c+f(x) are either ±∞, graphically, the function will have a vertical asymptote at x = c

• jump discontinuities occur at x = c when lim_x—c-f(x) ≠ lim_x—c+f(x), graphically, the function will jump from one value to another at x = c

to solve, factor denom and do a limit for each factor

definition of absolute value

• if x ≥ 0, |x| = x

• if x < 0, |x| = −x

delta-epsilon proofs

• if lim_x—af(x) = L, then given any number ε > 0, no matter how small, there is a number δ > 0 such that whenever 0 < Ix-aI < δ then If(x) - LI < ε

  • the goal is to choose a δ such that all of the above relationships are true. the main idea in the process is to find a relationship between x and δ so that you can show If(x)-LI is less than something in terms of δ and constants. once you have shown that, set the expression with the δ and constants equal to ε and solve for δ. the expression you get for δ is the δ you were originally supposed to choose

  • while solving, you will need to use the relationship that Ix-aI < δ to convert from terms of x to terms of δ

  • for quadratic and cubic functions, you will often have to put an extra restriction on the value of δ (usually you will want to make Ix-aI < 1 unless the value of a is 1. this requires your answer to end up as an expression δ = min{1, z} where z is the final expression you get in terms of ε

  • for cubic functions, you will often use the triangle inequality, Ia + b + cI ≤ IaI + IbI + IcI to help break down your expression and find a value of δ

• given ε > 0

• suppose 0 < |x-a| < δ

• then |f(x) - L| < ε

  • …x-a < …δ = ε/…ε = ε

• choose δ = …ε

delta-M proofs

• if lim_x—af(x) = ∞, then given any number M > 0, no matter how large, there is a number δ > 0 such that if Ix-aI < δ, then f(x) > M

• these types of proofs will generally be done for one-sided limits

• in the case that lim_x—af(x) = -∞, then given any number M < 0, no matter how large in absolute value there is a number δ > 0 such that if Ix-aI < δ, then f(x) < M

  • given M > 0 (M < 0 if -∞)

  • suppose 0 < |x-a| < δ

    • since x-a (one-sided), x-a </> 0 so 0 </> x-a </> δ and …x-a >/< …δ

  • then …x-a < …δ = M/…M = M

  • choose δ = …M

N-epsilon proofs

• if lim_x→∞f(x) = L, then given any number ε > 0, no matter how small, there is a number N > 0 such that if x > N > 0, then If(x)-LI < ε

• if lim_x→-∞f(x) = L, then given any number ε > 0, no matter how small, there is a number N < 0 such that if x < N < 0, then If(x)-LI < ε

  • when x—-∞, be especially careful with signs when dealing with your inequality

• given ε > 0

• suppose x > N > 0 (x < N < 0 if -∞)

  • since x > 0, …x > 0, so |f(x)| = f(x) and x > N

    • ..x > …N, …x < ..N

• then |f(x) - L| = f(x) - L < …N = ε

  • ..N = ε/…ε = ε

• choose N = …ε

N-M proofs

• if lim_x—∞f(x) = ∞, then given any number M > 0, no matter how large, there is a number N > 0 such that if x > N > 0, then f(x) > M

• if lim_x—∞f(x) = −∞, then given any number M < 0, no matter how large in absolute value, there is a number N > 0 such that if x > N > 0, then f(x) < M

• if lim_x—−∞f(x) = ∞, then given any number M > 0, no matter how large, there is a number N < 0 such that if x < N < 0, then f(x) > M

• if lim_x—−∞f(x) = −∞, then given any number M < 0, no matter how large in absolute value, there is a number N < 0 such that if x < N < 0, then f(x) < M

• often, it will be necessary to add an extra restriction on the value of N for these problems so that you can manipulate an expression similar to x > 1 so that it looks like a portion of the numerator or denominator of the original function by adding constants or values of x to both sides of the inequality

• also, it can be useful to turn the numerator and denominator of your original function into monomials so that you can cancel an x from both

• given M > 0 (M < 0 x— −∞)

• suppose x > N > 0 (x < N < 0 x— −∞)

• then f(x) < …. < …x > …N = …M

• choose N = …M

methods to know

• pascal’s triangle

  • 1

  • 1 1

  • 1 2 1

  • 1 3 31

  • 1 4 6 4 1

  • 1 5 10 10 5 1

  • 1 6 15 20 15 6 1

• triangle inequality theorem

  • | a + b | <= | a | + | b |