Step 1: Look at the first limit $k$. We must evaluate $k=\displaystyle\lim_{x\to0}\dfrac{|x|}{\sqrt{x^4+4x^2+5}}$. Step 2: Substitute the limiting behaviour. As $x\to0$, the numerator $|x|\to0$ while the denominator $\sqrt{x^4+4x^2+5}\to\sqrt{5}$, which is nonzero. Step 3: Conclude $k$. A vanishing numerator over a nonzero limit gives $k=\dfrac{0}{\sqrt5}=0$. Step 4: Look at the second limit $l$. Here $l=\displaystyle\lim_{x\to0}x^4\sin\!\left(\dfrac{1}{3\sqrt{x}}\right)$. The sine is bounded between $-1$ and $1$. Step 5: Apply the squeeze theorem. Then $-x^4\le x^4\sin(\cdot)\le x^4$, and both bounds tend to $0$ as $x\to0$, so $l=0$. Step 6: Add the parts. Therefore $k+l=0+0=0$. \[ \boxed{0} \]