the question is

lim x->0 ( 3cos(2x) - sin(3x)/5x )

however, this don't meet "0/0"

comes out as 3(1) - 0 = 3

how should i solve this problem?

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- Jan 27th 2010, 10:26 AMhaebinparklimit
the question is

lim x->0 ( 3cos(2x) - sin(3x)/5x )

however, this don't meet "0/0"

comes out as 3(1) - 0 = 3

how should i solve this problem? - Jan 27th 2010, 10:41 AMMoo
Hello,

Well it's in the form constant/0, which tends to infinity (the signs depends... Since the numerator goes to 3, the sign depends on what happens in the denominator, namely +infinity if x goes to 0 from the left, and -infinity if x goes to 0 from the right) - Jan 27th 2010, 01:11 PMhaebinpark
so do i have to go with

x goes to 0 from the left/right?? - Jan 27th 2010, 02:03 PMhaebinpark
= lim x->0 3cos2x - lim x->0 sin3x/5x

= lim x->0 3cos2x - 3/5 lim x->0 sin3x/3x

what should i do next? - Jan 28th 2010, 03:48 AMhaebinpark
where am i doing it wrong?

can someone put me on the right track pleaes?! - Jan 28th 2010, 04:33 AMHallsofIvy
First, please write your problem correctly- you first appeared to be talking about (3 cos(2x)- sin(3x))/(5x), not (3cos(2x)- sin(3x)/5x) as your wrote.

If it is in fact the first of those, then, because the numerator goes to 3 and the denominator goes to 0, the limit**does not exist**.

But in your last post, you were back to 3 cos(2x)- sin(3x)/(5x). If that is correct then, yes, you can break it into two limits : $\displaystyle \lim_{x\to 0} 3 cos(2x)- \lim_{x\to 0} (3/5) sin(3x)/(3x)$.

The first, $\displaystyle \lim_{x\to 0} 3 cos(x)$ should be easy- cosine is continuous for all x. For you have, cleverly, multiplied both numerator and denominator by 3 to get the denominator the same as the argument of sine. I presume you did that because you know that $\displaystyle \lim_{a\to 0}\frac{sin(a)}{a}= 1$. Use that!