
first post
Hi, my name is Mike. I'm a student of Cal State Fullerton. I have this problem with L'hospital rule:
the Problem: find values of a and b that satisfy the following
lim (sin x/x^3 + a + b/x^2) = 0
x>0
I tried to combined the whole equation into one using common denominator but it's doesn't seem right at all because I have 2 unknown variables.

Re: first post
Hello and welcome, Mike!
I recommend in the future posting your problems in the appropriate forums, as they will receive more and prompt attention that way.
I think your first move is a good one, i.e., to write:
$\displaystyle \lim_{x\to0^{+}}\frac{\sin(x)+ax^3+bx}{x^3}=0$
We have the indeterminate form 0/0, so application of L'Hôpital's rule gives:
$\displaystyle \lim_{x\to0^{+}}\frac{\cos(x)+3ax^2+b}{3x^2}=0$
In order for this limit to be bounded, we require the numerator to tend to 0, so we need $\displaystyle b=1$, and we may write:
$\displaystyle \lim_{x\to0^{+}}\frac{\cos(x)+3ax^21}{3x^2}=0$
We have again the indeterminate form 0/0, so application of L'Hôpital's rule gives:
$\displaystyle \lim_{x\to0^{+}}\frac{\sin(x)+6ax}{6x}=0$
We have again the indeterminate form 0/0, so application of L'Hôpital's rule gives:
$\displaystyle \lim_{x\to0^{+}}\frac{\cos(x)+6a}{6}=0$
Can you now finish?