Originally Posted by
MarkFL2
$\displaystyle \lim_{x\to0^{+}}\frac{\sin(x)+ax^3+bx}{x^3}=0$
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:
$\displaystyle \lim_{x\to0^{+}}\frac{\cos(x)+3ax^2-1}{3x^2}=0$
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?