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Explain how apply squeeze principle to evaluate the limit cos^2 2x/(3-2x)? x->oo

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shoppana | Student, Undergraduate | eNoter

Posted December 2, 2011 at 1:04 AM via web

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Explain how apply squeeze principle to evaluate the limit cos^2 2x/(3-2x)?

x->oo

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted December 2, 2011 at 1:24 AM (Answer #1)

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Remember that the values of cosine functions are larger than -1 and smaller than 1, therefore the square of the values of cosine function are larger than zero and smaller than 1.

`0 =lt cos^2 (2x) =lt 1`

Because `x-gt+oo =gt 3-2x lt 0`

Divide the inequality by the negative amount 3-2x and reverse the direction of the inequality.

`0/(3-2x) gt= (cos^2 (2x))/(3-2x)gt= 1/(3-2x)`

Evaluate the limits:

`lim_(x->oo)` `0/(3-2x)`  `gt=`  `lim_(x-gtoo) ``(cos^2 (2x))/(3-2x)`  `gt=` `lim_(x-gtoo)``1/(3-2x)`

`0 gt= lim_(x-gtoo) (cos^2 (2x))/(3-2x) gt= 0`

Use the squeeze limit and conclude that `lim_(x-gtoo)` `(cos^2 (2x))/(3-2x) = 0`

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