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?
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)
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!