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

x->oo

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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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