Question

Use the squeeze theorem to determine:

LIM: x --> infinity [3-sin(e^x)]/[ V(X^2 + 2)

can be written as [3 minus Sine (exp x)] divided by [ x- squared plus 2]

Answer 1

Greetings, user ^_^
Welcome to OpenStudy.
Without further ado, allow me to explain the squeeze theorem (not in detail, though ;) )

\[\Large \lim_{x\rightarrow \infty}\frac{3-\sin(e^x)}{x^2+2}\]
First, however, is that^ the correct expression?

Answer 2

yes, it is the corerct expression

Answer 3

Well then, we consider this function...
\[\Large f(\color{red}x) = \frac{3-\sin(e^\color{red}x)}{\color{red}x^2 + 2}\]
And look for functions to 'squeeze' it with.

Can you think of a function that will always be less than this function, no matter what value x takes?

Answer 4

Hey, @MERTICH
stay with me :>

Answer 5

ok,

Answer 6

-1<sin(e^x<1 because i raised e to 100, and the sin is 0.999,

Answer 7

@terenzreignz hope you r still there

Answer 8

I could say the same thing about you... -_-

And I can see where you're going with this, so, let's just be a little more relevant...
\[\Large f(\color{red}x) = \frac{3\color{blue}{-\sin(e^\color{red}x)}}{\color{red}x^2 + 2}\]

What's the lowest value \(\Large -\sin(e^x)\) could possibly take?

Answer 9

I think its -1, I mean , sin x can always be between 1 and -1. correct me

Answer 10

That is correct ^_^Therefore, f(x) can never be less than ...
\[\Large f(\color{red}x) = \frac{3-\sin(e^\color{red}x)}{\color{red}x^2 + 2} \ge \frac{3\color{green}{-1}}{x^2+2}=\frac{2}{x^2+2}\]
Right? ^_^

Answer 11

so in other words as f(x) approaches infinity, it becomes 0. If you can please give a brief desription of what just happened there, and if possible, draw closer to the equation f(x)=h(x)=g(x).

Answer 12

Hold your horses... we haven't set it up yet, we have only ever found a lower bound for your function... what about an upper bound?

Answer 13

i think instead of 3-1 we put 3+1 on the numerator from -(-1), to make 4/( x^2 + 2) ??

Answer 14

still waiting for the upper bound, @terenzreignz

Answer 15

You already got it... I thought you could take it from there... Now...
\[\Large \frac{2}{x^2+2}\le \frac{3-\sin(e^x)}{x^2+2}\le\frac{4}{x^2+2}\]
Can you squeeze it now?

Answer 16

yes, i think. because the power on the denominator of the upper bound is greater, as x--> infintiy , it become zero, and the same applies to the lower bound equation, thus, when SQUEEZED, LIM X--> INFINITY, f(x) =0. thank you very much, you have given me more than an answer!

Answer 17

No problem ^_^

Promise me you'll practice, though :)

Answer 18

Yep, i am still new to the squeeze thoerem, And I will get there soon.