Question

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

Answer 1

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

Answer 2

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

Answer 3

or yea L'Hospital's Rule

Answer 4

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

Answer 5

the first one Turing thank you

Answer 6

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

Answer 7

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.

Answer 8

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

Answer 9

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