I'm going to guess the function is
[tex]f(x)=\begin{cases}\frac{9x-3\sin3x}{5x^3}&\text{for }x\neq0\\c&\text{for }x=0\end{cases}[/tex]
and that you are asked to find [tex]c[/tex] such that [tex]f(x)[/tex] is continuous at [tex]x=0[/tex]. For this to be the case, we require that the limit at [tex]x=0[/tex] exists and is equal to [tex]c[/tex]:
[tex]\displaystyle\lim_{x\to0}f(x)=f(0)=c[/tex]
L'Hopital's rule applies, since evaluating directly at [tex]x=0[/tex] yields an indeterminate form [tex]\dfrac00[/tex]:
[tex]\displaystyle\lim_{x\to0}\frac{9x-3\sin3x}{5x^3}=\lim_{x\to0}\frac{9-9\cos3x}{15x^2}[/tex]
Taking [tex]x=0[/tex] again yields [tex]\dfrac00[/tex]. Applying once more, we have
[tex]\displaystyle\lim_{x\to0}\frac{27\sin3x}{30x}[/tex]
We could use the rule again, or make use of the fact that
[tex]\displaystyle\lim_{x\to0}\frac{\sin ax}{ax}=1[/tex]
for any non-zero value of [tex]a[/tex]. So
[tex]\displaystyle\lim_{x\to0}\frac{27\sin3x}{30x}=\frac{27}{10}\lim_{x\to0}\frac{\sin3x}{3x}=\dfrac{27}{10}=c[/tex]
which agrees with the result we would have gotten had we used L'Hopital's rule:
[tex]\displaystyle\lim_{x\to0}\frac{27\sin3x}{30x}=\lim_{x\to0}\frac{81\cos3x}{30}=\dfrac{81}{30}=\dfrac{27}{10}[/tex]
