Evaluate a trigonometric limit using L'Hopital's Rule

Substituting $x=0$ gives $\sin(0)/\tan(0) = 0/0$, which is an indeterminate form, so L'Hopital's Rule applies. Differentiate the numerator and denominator separately: the derivative of $\sin(3x)$ is $3\cos(3x)$, and the derivative of $\tan(5x)$ is $5\sec^2(5x)$.

Now evaluate $\lim_{x\to 0} \dfrac{3\cos(3x)}{5\sec^2(5x)}$ by direct substitution, since this new expression is no longer indeterminate. At $x=0$, $\cos(0)=1$ and $\sec^2(0)=1$, so the limit equals $\dfrac{3(1)}{5(1)} = \dfrac{3}{5}$.