The limit of that, for x going to -5, is -5- 3= -8.
For x< -5.
∣x+5∣=−(x+5) so this is
−(x+5)(x+5)(x−3)=−(x−3). The limit of that, for x going to -5, is
−(−5−8)=8.
The limit does not exist but because the two one-sided limits are not the same, not because the denominator is 0. If the problem had been
x→−5limx+5x3+2x−15, without the absolute value, the limit would be -8 even though the denominator still goes to 0.