Squeeze Law

[confused_07]

Can't seem to understand this concept. Homework question is:

Use the squeeze law of limits to find the limit. limit(x->0) x^2 sin^2 10x

 

[pka]

Can't seem to understand this concept. Homework question is: Use the squeeze law of limits to find the limit. limit(x->0) x^2 sin^2 10x

There is absolutely no reason to use ‘the squeeze law of limits’ for this!
The limit is zero without any further effort.

 

[confused_07]

That's what I thought. I have to show my work, so this is what I got:

-x^2 < x^2 sin^2 10x < x^2

f[x] = -x^2
= lim (x->0) = -(0)^2
= 0

h[x] = x^2
= lim (x->0) = 0^2
= 0

 

[pka]

There is a theorem that states the limit of a product is the product of the limits. That is absolutely fundamental as opposed to ‘the squeeze law of limits’. If you had \(\displaystyle x^2 \sin ^2 \left( {\frac{1}{x}} \right)\) then we certainly need ‘the squeeze law of limits’ to evaluate the limit as x approaches zero.

 

[confused_07]

I am in Chapter 2 of Calculus, and it has half a page that describes the squeeze law. All I get from it is that as two functions come together at a point, then the third function will equal that point as well.

Then I get this homework question to eval the limit using this Law.

Thanks for the help. After using the Product Law, I understand it a little bit more.

 

[soroban]

Hello, confused_07!

Use the Squeeze Law of limits to find: \(\displaystyle \L\;\lim_{x\to0} x^2\cdot\sin^210x\)

They expect you to find two functions to do the "squeezing".
. . It's an exercise in Imagination.

That is, find a function that is always less than \(\displaystyle f(x)\) which has a limit
. . and one that is always greater than \(\displaystyle f(x)\) whose limit is the same.
Then \(\displaystyle f(x)\) will be "squeezed" to that limit.


Since \(\displaystyle \left(x\cdot\sin10x)^2\) is never negative,
. . we have: \(\displaystyle \,0 \:\leq\:x^2\cdot\sin^210x\)

Since \(\displaystyle \sin^210x\:\leq\:1\), multiply both sides by \(\displaystyle x^2.\)
. . we have:\(\displaystyle \,x^2\cdot\sin^210x \:\leq\:x^2\)

Hence, we have: \(\displaystyle \:0 \:\leq \:x^2\cdot\sin^210x\:\leq \:x^2\)


\(\displaystyle \text{Take the limit: }\L\:\underbrace{\lim_{x\to0} 0}_{\downarrow_0} \;\leq \;\lim_{x\to0}x^2\cdot\sin^210x \:\leq \:\underbrace{\lim_{x\to\0}x^2}_{\downarrow_0}\)

\(\displaystyle \text{Therefore: }\L\:\lim_{x\to0}x^2\cdot\sin^210x\;=\;0\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Think of a trash compactor.

There is an apple somewhere between the two movable walls.

The two walls are approaching a central position; call it \(\displaystyle 0.\)

As long as the apple remains between the two walls,
. . the apple will end up at position \(\displaystyle 0\;\) (applesauce!)

 

[Count Iblis]

Lim x --> a f(x) = f(a) is f is continuous at a. So you already know that the limit is zero in this case.