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#1 2013-12-19 00:42:00

MERTICH
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Squeeze Theorem

Use the squeeze theorem to determine

                                                                    LIM                                [ 3 -sin(e^x)]/                 3 minus Sin(exp x) divided by
                                                                   x--> infinity                     [V (X^2 + 2)]                   square root of ( x squared + 2)

#2 2013-12-19 00:48:31

MERTICH
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Re: Squeeze Theorem

I would also appreciate a stage by stage go...

#3 2013-12-19 00:55:05

zetafunc.
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Re: Squeeze Theorem

Use the fact that |sin(ex)| ≤ 1 and play with the inequality until you can bound your function.

#4 2013-12-19 01:03:29

MERTICH
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Re: Squeeze Theorem

@ zetfunc , you mean like, since I know that no matter what, -1<sinx < 1 , so I should subsititute sin(e^x) with 1 and -1, in the expression to get my lower and upper bound?

Last edited by MERTICH (2013-12-19 01:10:09)

#5 2013-12-19 01:11:44

zetafunc.
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Re: Squeeze Theorem

The sine function, regardless of the argument (and as long as it is real), is always at least bounded between -1 and 1.







Do you see what we are trying to do?

#6 2013-12-19 01:29:37

MERTICH
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Re: Squeeze Theorem

yes, but then how do you fit in the rest of the equation, or is it 2/( x^2 +2) </= f(x) </= 4/ (x^2 +2)  ?

#7 2013-12-19 01:44:24

anonimnystefy
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Re: Squeeze Theorem

Yes, that is correct. The limit of the left and right bound are easy to get, so you should be able to get the original limit now.


The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

#8 2013-12-19 02:08:23

zetafunc.
Guest

Re: Squeeze Theorem

Yes (with the square root signs).

Now you just need to show that the limits of both bounds go to zero as x approaches infinity.

#9 2013-12-19 03:02:47

MERTICH
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Re: Squeeze Theorem

that is to say the denominator power of both bounds is greater than the numerator power, such that when x approaches infinity, f(x) becomes zero.....thank you guys!

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