lim(x-16) (4-sqrt(x))/(16-x^2) ?? I got 0/240... but apparently that's incorrect.

Question

lim(x->16) (4-sqrt(x))/(16-x^2) ?? I got 0/240... but apparently that's incorrect.

anonymous · Answer

no it becomes 0/0 so you have to do some algebra first

rulnick · Answer

Is it (16-x)^2 in the denominator? You know L'Hopital?

anonymous · Answer

or yea L'Hospital's Rule

turingtest · Answer

is the problem$$\lim_{x\to16}\frac{4-\sqrt x}{16-x^2}$$or$$\lim_{x\to16}\frac{4-\sqrt x}{(16-x)^2}$$?

anonymous · Answer

the first one Turing thank you

anonymous · Answer

oh pellet no it's actually 16x-x^2 that's the denominator whoops copied it wrong

rulnick · Answer

Hi, calyne.  Then what you want is
numerator: lim as x to 16 of (-1/2) x^(-1/2)
denominator: lim as x to 16 of (16-2x)
by L'Hopital.

rulnick · Answer

The result is (-1/8) / (-16) = 1/128.

anonymous · Answer

yeah i got it thank you sorry openstudy was buggin didn't let me post that i got the answer i appreciate it though