Question

ds

Answer 1

You don't have to do this step,
but it might make things easier.
I would recommend making a substitution.
\(\large\rm \dfrac{1}{t}=u\).

Then notice that as \(\large\rm t\to0^-\), we have \(\large\rm u\to -\infty\)
I can explain that further if it's confusing ^.
So our limit becomes:\[\large\rm \lim_{t\to0^-}\sin^2\left(\frac{1}{t}\right)3^{1/t}=\lim_{u\to-\infty}\sin^2(u)3^u\]And then Squeeze Theorem is a little easier from this point.

Answer 2

I don't think my teacher wants me to make the substitution but for me it makes it easier so far!

Answer 3

Either way is fine, let's stick with the original then.

With Squeeze Theorem,
you want to start with only your "troublesome piece" and try to get some bounds on it.
In this problem, the sine is what's causing trouble for us.
It fluctuates back and forth forever as t gets closer to 0.

So let's put some hard boundaries on sine and continue from there.

Answer 4

ok

Answer 5

Well we know that sine is stuck between -1 and 1.
But let's just right to sine squared if we can.
That gets rid of all of the negative values,\[\large\rm 0\le \sin^2\left(\frac{1}{t}\right)\le 1\]

Answer 6

Multiply all sides by 3^(1/t),\[\large\rm 0\cdot3^{1/t}\le \sin^2\left(\frac{1}{t}\right)3^{1/t}\le 1\cdot3^{1/t}\]\[\large\rm 0\le \sin^2\left(\frac{1}{t}\right)3^{1/t}\le 3^{1/t}\]

Answer 7

From here,
if you can show that the left and right most sides of this inequality converge to 0,
then the stuff in the middle has to converge to 0 by the Squeeze Theorem.

The left-most side is pretty straight-forward, ya?\[\large\rm \lim_{t\to0^-}0=0\]

Answer 8

For the right-most side,
set up the limit and do something to show it's zero.
Maybe that u-sub if it makes it easier.

Answer 9

how do i show that the right side is 0 though without usub?

Answer 10

left side makes sense though haha

Answer 11

this whole sandwich theorem is all new to me so im still trying to figure everything out haha

Answer 12

Sandwiches are delicious :o so don't let it scare you.
We're just trying to show that as t approaches 0 from the left,
our inequality is approaching:
\(\large\rm 0\le stuff\le 0\)
The stuff is sandwiched between 0 and 0, so it has to be 0.

Answer 13

Without u-sub? Ummm

Answer 14

Probably with some simple words to justify it... like umm

Answer 15

its just that we haven't been taught it so i don't want him to get upset or anything

Answer 16

maybe as t gets closer to 0, the right side approaches 0? so since there both =0 the original limit is 0?

Answer 17

You're just restating that the right limit is 0
not "showing" it yet lol

but that's ok, it's hard to say how thorough your teacher wants you to be.
Hopefully you can make some sense out of why it should be zero though.\[\Large\rm \lim_{t\to0^-}3^{1/t}\]Let's plug a number really close to zero in, that's below zero,
just to get an idea of what is going on,\[\Huge\rm 3^{\frac{1}{\color{orangered}{t}}}\approx3^{\frac{1}{\color{orangered}{-\frac{1}{99999}}}}\]We're dividing by a fraction in the exponent, so we can flip it,\[\large\rm =3^{-99999}\]Rule of exponents lets us write it like this,\[\large\rm \frac{1}{3^{99999}}\]Which is a really really small number, almost 0.

Answer 18

I could have chosen a decimal I suppose like t=-0.0000001,
but fractions are easier to flip.

Answer 19

But ya maybe just use some words and teach will be ok with it :)

Since \(\large\rm 0\to0\) and \(\large\rm 3^{1/t}\to0\) as \(\large\rm t\to0^-\),
by the Squeeze Theorem we can conclude that
\(\large\rm \sin^2\left(\frac{1}{t}\right)3^{1/t}\to0\) as well.

Answer 20

ok i see what your doing here

Answer 21

thanks!

Answer 22

reagarding the bounds, why is it 0 and 1?

Answer 23

Are you ok with the normal boundaries of the sine function?
Like if I wrote this\[\large\rm -1\le \sin(x)\le 1\]does it make sense?

Answer 24

ya

Answer 25

sin^2 doesn't though...

Answer 26

Well, when you square something, it eliminates any negative values, ya?