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- anonymous

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- schrodinger

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- anonymous

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- zepdrix

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.

- anonymous

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

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- zepdrix

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.

- anonymous

ok

- zepdrix

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,\[\large\rm 0\le \sin^2\left(\frac{1}{t}\right)\le 1\]

- zepdrix

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

- zepdrix

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?\[\large\rm \lim_{t\to0^-}0=0\]

- zepdrix

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.

- anonymous

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

- anonymous

left side makes sense though haha

- anonymous

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

- zepdrix

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.

- zepdrix

Without u-sub? Ummm

- zepdrix

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

- anonymous

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

- anonymous

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

- zepdrix

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.

- zepdrix

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

- zepdrix

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.

- anonymous

ok i see what your doing here

- anonymous

thanks!

- anonymous

reagarding the bounds, why is it 0 and 1?

- zepdrix

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?

- anonymous

ya

- anonymous

sin^2 doesn't though...

- zepdrix

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

- zepdrix

|dw:1444368409918:dw|

- zepdrix

|dw:1444368473226:dw|

- zepdrix

It can't give us anything negative, so we lose half of that range from the sine function.

- anonymous

ok makes sense! thanks

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