limit of a function general idea
a function f(x) has a limit L at x = a if f(x) is close to L whenever x is close to a
if we have an acceptable error (ε >0) between f(x) + L, need |f(x) -L| < ε for all x in some interval around a
if for any ε >0 i can find a distance d >0 such that
|f(x) - L| < ε for all 0<|x-a|<d
then f(x) has a limit L as x tends to a
limit definition
f(x) has a limit L as x tends to a if for all ε >0 there exists d >0, such that |f(x) -L| < ε for all x such that 0 < |x-a| < d
we write lim f(x) as x -> a = L or f(x) -> L as x-> a
if there is no L the limit does not exist
continuity
a function f(x) is continuous at x = a if:
- f(a) exists
- lim f(x) as x->a exists
- lim f(x) as x->a = f(a)
function continuous on a subset S of its domain
if its continuous at every point in S
continuous if its continuous at every point in its domain
right sided limit
L + = lim f(x) as x tends to a+ (tends to a from above) if for all ε >0 there exists d >0, such that |f(x) -L+| < ε for all x such that 0 < x-a< d
x> a
left sided limit
L- = lim f(x) as x tends to a- (tends to a from below) if for all ε >0 there exists d >0, such that |f(x) -L-| < ε for all x such that 0 < a-x < d
x<a
limit exists if
if and only if L+ and L- both exist and are =
L = L+ = L-
3 types of discontinuity
removable discontinuity
L exists but f(a) =/= L
we can always remove this to make a continuous function
g(x) is
f(x) if x =/=a or L if x = a
jump discontinuity
L + and L- exists but L+ =/= L-
(so L does not exist)
infinite (or essential) discontinuity
at least one of L+ or L- does not exist
facts about limits
facts about continuity
pinching / squeezing theorem
If g(x) <= f(x) <= h(x) for x =/= a in some open interval containing a and if lim g(x) = lim h(x) = L
then lim f(x) = L
2 trig limits
lim sinx/x = 1 as x -> 0
lim (1-cos x)/x = 0 as x -> 0
limits to infinity rough idea
a function as a limit to infinity if f(x) can be kept arbitrarily close to L by making x sufficiently large
limit to infinity
f(x) has a limit L as x -> ♾️ if for all epsilon >0 there exists S >0 such that |f(x) - L < epsilon for all x >S
(same for -♾️ but x<S)
method for limit to infinity
make substitution u = 1/x then tend u -> 0+ for f(1/u)
asymptote and limits
the graph of f(x) has a horizontal asymptote to the right (or left) at y = L if the lim x->♾️ f(x) = L (or lim x -> -♾️ f(x) = L)
intermediate value theorem (IVT)
if f(x) is continuous on [a,b] and u is any value between f(a) and f(b)
then there exists c ∈ (a,b) such that f(c) = u
sgn(x)
1 for x>0
0 for x = 0
-1 for x<0
application of IVT
is finding roots of a function
if f is continuous on [a,b] and f(a) <0 <f(b) or f(b)<0 <f(a) then by IVT theres at least one root such that f(x) = 0 between a and b