Discuss the limiting behaviour of abs(x^2 - 9)/(x-3), as x-3 from the negative side, positive side, and as x-a.

Question

Discuss the limiting behaviour of abs(x^2 - 9)/(x-3), as x->3 from the negative side, positive side, and as x->a.

anonymous · Answer

$$\left| x^2-9 \right|/(x-3)$$

I know how to work out regular rational functions, but the absolute value has me stumped.
Can someone point me in the right direction?

hartnn · Answer

since numerator is ALWAYS positive(because of mod),this function will be negative for x<3(from denominator)....and will be positive for x>3,as x->3,it will be just x+3=6,function will tend to 6....if there was no mod sign,then function would be x+3 and would go negative from x<-3....

anonymous · Answer

Thanks. I don't quite understand the second part though. Why is the function rediced to just x+3=6 for x>3?

hartnn · Answer

its for x->3...x tends to 3......then u can cancel out x-3 from numerator and denominator

anonymous · Answer

Ah, so you can ignore the absolute value part for x>3, then you have $$\frac{(x-3)(x+3)}{(x-3)}$$, which cancels out to just x+3. Is that correct?

hartnn · Answer

u can ignore for x-> 3(x tends to 3),not for x>3

anonymous · Answer

OK, I think I get it. For this function though,

$$\frac{\left| x-5 \right|}{x-5}$$

how do you evaluate x->5 from the negative side and the positive side without getting 0?