calculus sin frontera
Section 1.x: Weird Limits
Level II Alert This one goes out to all those strange limits that noone loves. We’ll start with limx→∞ sin( x ). A little graph shows sine
doesn’t look remotely like it has a limit – it wobbles between +1 and
-1 and never settles down.
But Let’s try it: let’s pretend I believe limx→∞ sin( x ) = 1. What’s
gonna happen? I say I believe that sine settles down near one. But
the very next cycle, sine hits zero: sine refuses to stay close to just one
number.
Figure 1: sin( x )
Looking like it’ll wobble forever.
Now let’s make that intuition into math (algebra, actually): I’m supposed to believe that there’s a number n, and when x > 10n , sin( x )
and 1 agree to, say, two decimal places. Since sine can’t be bigger
than 1, sin( x ) would have to be between 1 and, say, .95: for x > 10n ,
.95 ≤ sin( x ) ≤ 1. That means if I take x = 10n π, then I certainly have
x > 10n . So I’d have to believe that .95 ≤ sin(10n π ) ≤ 1. But sine is
zero at all multiples of π, and if I use that, .95 ≤ 0 ≤ 1. So not gonna
happen.
My next limit is a variant: limx→∞ sin( x )/x. The idea here is that the
sine wobbles, but the 1/x goes to zero, so that multiplying the 1/x
into sine will tame the wobble. Like in Figure 2: the wobble is tamed.
Figure 2: sin( x )/x
Taming the wobble.
Here it is in math: The wobble is
−1 ≤ sin( x ) ≤ 1
And the taming is
−
1
sin( x )
1
≤
≤
x
x
x
Now for the limit:
lim −
x →∞
1
sin( x )
1
≤ lim
≤ lim
x →∞
x →∞ x
x
x
And now
0 ≤ lim
x →∞
Figure 3: A Close Up
Here I can see sine being squashed
down by the green ±1/x.
sin( x )
≤0
x
As promised, all the wiggle has been squashed out of sine; it has no
choice but to go to zero.
A Neat Trick This trick interchanges zero and infinity. It’s easy: if I
let y = 1/x, then as x → ∞, y → 0. I’ll use this to change variables in
limy→0 sin( 1y ): the limy→0 ↔ limx→∞ ; the sin( 1y ) ↔ sin( x ), and so:
1
lim sin( ) = lim sin( x ) = does not exist
x →∞
y
y →0
Figure 4: Flipping The Sine
The wiggles of sine are compressed
near zero.
1
2
kathy davis
I get two limits for the work of one. All the wiggling I saw in Figure
2 as x went to infinity, is switched over to zero, and compressed
down; as in Figure 4.
And, let’s do it again, with limy→0 y sin( 1y ). The only term that’s
different is the factor of y in front of the sine: since y = 1/x, we get
1
sin( x )
y sin( ) ↔
y
x
and then
1
sin( x )
lim y sin( ) = lim
=0
x →∞
y
x
y →0
Figure 5: Squashing The Sine, II
The wiggles of sine near zero are
squashed between the two green lines.
You can see the squashing in Figure 5.
This trick, switching zero and infinity, woks for other functions too.
In class, we showed limx→∞ ln( x ) = ∞. Now I’m going to switch that
infinity to zero: let x = 1/y. Then,
lim ↔ lim
y →0+
x →∞
1
ln( x ) ↔ ln( ) = − ln(y)
y
lim ln( x ) = ∞ ↔ lim − ln(y) = ∞
x →∞
y →0+
So,
lim ln(y) = −∞
y →0+
This means that the log function will have a vertical asymptote at at
x = 0+ , as in Figure 6.
Level III Alert This trick of changing one variable into another
is something we’ll use repeatedly in the course, in the chapters on
integration. But – here’s an example you might not expect:
√
1− 3 x
0
lim
=
0
x →1 x − 1
Since it’s a 0/0, I want to factor out an x − 1 from the numerator. If
The numerator had a square root instead of a cube root, I’d multiply
by the conjugate. But that’s the conjugate for a cube root?
Figure 6: The Old Switcheroo
When I switch the limit as x → ∞ of
ln( x ), I create a vertical asymptote for
log at x = 0+ .
calculus sin frontera
Since I don’t know the answer to that (secretly I do [show-off!]), I’ll
just get rid of the cube root: let x = y3 . Then:
lim ↔ lim
x →1
y →1
√
1− 3 x
1−y
↔ 3
x−1
y −1
Difference of cubes, I can handle:
1−y
−1( y − 1)
−1
=
= 2
y3 − 1
(y − 1)(y2 + y + 1)
y +y+1
and,
√
1− 3 x
−1
1
=−
lim
= lim 2
3
x →1 x − 1
y →1 y + y + 1
By the way, if you take the 1/(y2 + y + 1) and change back to x, you
get
1
2
1
x3 + x3 + 1
and this tells you that the conjugate of 1 −
:) :) Sleep tight ...
√
3
2
1
x is x 3 + x 3 + 1
3