#### To determine

**To compute:** limx→a2a3x−x4−aaaxa−ax3b

#### Answer

limx→a2a3x−x4−aaaxa−ax3b=169a

#### Explanation

limx→a2a3x−x4−aaaxa−ax34=limx→a2a3x−x4−aa2xa−ax34

Applying l’Hospital’s Rule, we get

limx→a2a3x−x4−aa2x3a−ax3b=limx→a122a3x−x4×(2a3−4x3)−a.13(a2x)−23×a2−14(ax3)−34.3ax2=limx→a2a3−4x322a3x−x4−a.13(a2x)−23×a2−14(ax3)−34.3ax2Substitute the value x=a=2a3−4a322a3a−a4−a.13(a3)−23×a2−14(aa3)−34.3aa2=−a−a3−34=169a

**Conclusion:** The value of the limit is limx→a2a3x−x4−aaaxa−ax3b=169a