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# Impossible limits question watch

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1. find the limit as x tends to 0 of (cos^2 x-2)/sinhx+1
2. Is it cos^2(x-2) or cos^2(x) - 2

?

In any case, can't you just plug x = 0 into the expression?
3. Sorry if anyone saw my previous post - I misread the question.

Like tazarooni said, cos and sinh both have limits as x tends to 0 so by the algebra of limits, plugging x=0 will work.
4. well the questions is
limit as x tends to 0 of:- (cos^2 x)-2
sinhx+2

Basically when i applyl'hopital's rule, and differentiate, the top or bottom of the fraction is always zero, while the other is 1. Therefore I can see no way of getting a real answer, unless the answer is 0/1, as 1/0 isn't defined.
5. This isn't a suitable situation to use L'Hopital's rule (check the conditions!)

The numerator converges to -1, the denominator converges to 2. Hence the limit you want to find is -1/2.

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Updated: November 16, 2008
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