Math Is Fun Forum

MERTICH

Member

Registered: 2013-12-01

Posts: 18

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)

MERTICH

Member

Registered: 2013-12-01

Posts: 18

Re: Squeeze Theorem

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

zetafunc.

Guest

Re: Squeeze Theorem

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

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

zetafunc.

Guest

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?

MERTICH

Member

Registered: 2013-12-01

Posts: 18

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)  ?

anonimnystefy

Real Member

From: Harlan's World

Registered: 2011-05-23

Posts: 16,049

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.

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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.

MERTICH

Member

Registered: 2013-12-01

Posts: 18

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!