What, precisely, does it take to make a function continuous at x = c?
f continuous at c if continuous from left and from right and left limit = right limit
limx→cf(x) = f(c)
3 conditions
f(c) exists
limx→cf(x) exists
limx→cf(x) = f(c)
Continuous on an interval?
Continuous on interval [a,b] if
Right-continuous at a
Left-continuous at b
Continuous at all other points in [a,b]
Domain includes [a,b]
Graph of f(x) over [a,b] has no breaks
Example: hyperbola that is continuous on [0,n] but discontinuous on [-1,1]
What does the Intermediate Value Theorem say?
Intermediate value property = continuous function that = a
at one point and equals b at another takes on all values
between a and b between those points
Continuous function f(x) on [a,b], horizontal line between
f(a) and f(b) crosses graph of f
What does it mean to talk about limx→±∞f(x)?
Behavior as x gets arbitrarily big or arbitrarily small
What does it mean to say that limx→cf(x) = ±∞?
f(x) has vertical asymptote at c
f(x) becomes arbitrarily large (or small) as x approaches c
What is a 1-sided limit?
Limit as x approaches c but from only above or below
What is an asymptote?
Value (line on graph) that function approaches as x approches ±∞ but never actually reaches
b is (horizontal) asymptote for f(x) if limx→±∞f(x) = b
Can also have vertical asymptotes where f(x) approaches ±∞
Examples
Suppose I leave Geneseo at 8:00 some morning, driving to New
York City. At the same time Prof. X leaves Toronto driving to
Geneseo. Is there ever a time when I and Prof. X are exactly
the same distance from Geneseo (although in different places)?
Does the answer change if one or both of us stop, take detours,
backtrack, etc?
Yes, there has to be a time at which we’re the same
distance from Geneseo by intermediate value theorem
No, backtracking etc doesn’t matter
When can you “push” a limit inside a function application, i.e.,
when is limx→cf( g(x) ) = f( limx→cg(x) )?
Book theorem 10—when f is continuous at g(c)
Can you find
limx→π/4cos( (x2-π2/16) / (x-π/4) )? How about
limx→π/4tan( (x2-π2/16) / (x-π/4) )?
limx→π/4cos( (x2-π2/16) / (x-π/4) )
= cos( limx→π/4(x2-π2/16) / (x-π/4) )
= cos( limx→π/4(x-π/4)(x+π/4) / (x-π/4) )
= cos( limx→π/4(x+π/4)
= cos( π/2 )
= 0
The last few steps of this derivation as a drawing, if that’s more readable:
Is limx→π/2tan x defined? (No, left and right limits differ.)
What about limx→π/2+tan x? (Yes, tangent approaches
-∞ for x > π/2.)