Evaluate this limit, if it exists

[ppkjref]

Homework Statement

Evaluate the limit, if it exists. the limit of ((2/(3+h))^2-(4/9))/h as h approaches 0.

The Attempt at a Solution

So far I got,

[(2/(3+h))^2-(4/9)]/h
= [(2/(3+h))(2/(3+h))-(4/9)]/h

It seems nothing can cancel out when I find a common denominator.

 

[Dick]

Keep going. You haven't done anything yet. Find a common denominator in the numerator and simplify it. Do some algebra.

 

[supratim1]

use L'Hospital rule

 

[supratim1]

note that its 0/0 form

 

[Dick]

use L'Hospital rule

That would work. But this looks like a difference quotient from a derivative definition. Probably wouldn't be appropriate to use l'Hopital.

 

[ppkjref]

i noted that it is 0/0 form so therefore I am likely to get a real number.
[(2/(3+h))^2-(4/9)]/h
= [(2/(3+h))(2/(3+h))-(4/9)]/h
=[(4/(9+6h+h^2))-(4/9)]/h
=[(36-(36+24h+4h^2))/(9(9+6h+h^2))]/h
=[(-4h(h+6))/((3h+9)(3h+9))]/h
=[(-4h^2(h+6))/((3h+9)(3h+9))]

So the denominator is 81 and the numerator is 0?

 

[supratim1]

That would work. But this looks like a difference quotient from a derivative definition. Probably wouldn't be appropriate to use l'Hopital.

yes agreed. if that's the case, then it can be solved by doing some algebra, would just be a little more lengthy. only if it is required not to be done by L'Hospital.

 

[Dick]

i noted that it is 0/0 form so therefore I am likely to get a real number.
[(2/(3+h))^2-(4/9)]/h
= [(2/(3+h))(2/(3+h))-(4/9)]/h
=[(4/(9+6h+h^2))-(4/9)]/h
=[(36-(36+24h+4h^2))/(9(9+6h+h^2))]/h
=[(-4h(h+6))/((3h+9)(3h+9))]/h
=[(-4h^2(h+6))/((3h+9)(3h+9))]

So the denominator is 81 and the numerator is 0?

That's making my eyes hurt. Sorry. I think you were almost there in the fourth line. The 36's cancel. So you've got ((-24h+stuff)/(81+stuff))/h. Then a bad thing happened. How did the h pop up from the denominator into the numerator? It should have cancelled.

 

[ppkjref]

wow that's embarrassing. i'm an idiot hah