**Problem Statement:**Find the limit.(Use symbolic notation and fractions where needed. Enter DNE if the limit does not exist.)\[\lim_{(x,y) \to (0,0)} \frac{\sin(6xy)}{3xy} = \underline{\hspace{2cm}}\]**Instructions:**- The task is to evaluate the given limit of a function involving trigonometric and polynomial expressions as \((x, y)\) approaches \((0, 0)\).- The limit expression uses the common limit property that \(\lim_{u \to 0} \frac{\sin(u)}{u} = 1\).- If the limit does not exist, you should enter "DNE" in the answer box.

To evaluate the below limit. limx,y→0,0sin6xy3xyWe have to evaluate the limit at the origin. Let approach the origin using the below equation. y = mx Substitute y = mx in the above limit. It gives, limx,y→0,0sin6xy3xy=limx,y→0,0sin6x×mx3xmxlimx,y→0,0sin6xy3xy=limx→0sin6mx23mx2Apply L'Hopital rule: limx,y→0,0sin6xy3xy=limx→0ddxsin6mx2ddx3mx2limx,y→0,0sin6xy3xy=limx→0cos6mx2ddx6mx26mxlimx,y→0,0sin6xy3xy=limx→0cos6mx212mx6mxlimx,y→0,0sin6xy3xy=2limx→0cos6mx2limx,y→0,0sin6xy3xy=2limx→0cos6mx2Substitute x = 0 in it limx,y→0,0sin6xy3xy=2cos6m02limx,y→0,0sin6xy3xy=2cos0limx,y→0,0sin6xy3xy=2×1limx,y→0,0sin6xy3xy=2 Thus, the limit exist and equals to 2.limx,y→0,0sin6xy3xy=2