# limit

• January 27th 2010, 10:26 AM
haebinpark
limit
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?
• January 27th 2010, 10:41 AM
Moo
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)
• January 27th 2010, 01:11 PM
haebinpark
so do i have to go with
x goes to 0 from the left/right??
• January 27th 2010, 02:03 PM
haebinpark
= 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?
• January 28th 2010, 03:48 AM
haebinpark
where am i doing it wrong?
can someone put me on the right track pleaes?!
• January 28th 2010, 04:33 AM
HallsofIvy
Quote:

Originally Posted by haebinpark
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?

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 : $\lim_{x\to 0} 3 cos(2x)- \lim_{x\to 0} (3/5) sin(3x)/(3x)$.

The first, $\lim_{x\to 0} 3 cos(x)$ 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 $\lim_{a\to 0}\frac{sin(a)}{a}= 1$. Use that!