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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: Harlan's World
- Registered: 2011-05-23
- Posts: 16,016

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.

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

The knowledge of some things as a function of age is a delta function.

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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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