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)
I would also appreciate a stage by stage go...
Use the fact that |sin(e[sup]x[/sup])| ≤ 1 and play with the inequality until you can bound your function.
@ 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)
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?
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) ?
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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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.