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**MERTICH****Member**- Registered: 2013-12-01
- Posts: 18

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)

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**MERTICH****Member**- Registered: 2013-12-01
- Posts: 18

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

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**zetafunc.****Guest**

Use the fact that |sin(e[sup]x[/sup])| ≤ 1 and play with the inequality until you can bound your function.

**MERTICH****Member**- Registered: 2013-12-01
- Posts: 18

@ 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 (2013-12-18 02:10:09)*

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**zetafunc.****Guest**

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?

**MERTICH****Member**- Registered: 2013-12-01
- Posts: 18

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) ?

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,796

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.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**zetafunc.****Guest**

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.

**MERTICH****Member**- Registered: 2013-12-01
- Posts: 18

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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