lim (s-->16) of (4-s^1/2)/(s-16)
The limit as s approaches 16 of 4 minus square root s over s minus 16.
2007-09-25 13:34:06 · 5 answers · asked by Mrs.Harbi 3 in Science & Mathematics ➔ Mathematics
f(s) = -s^1/2
f'(s) = -1 / (2sqrts)
s = 16
f'(s) = -1 / 8
Answer: -1/8
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Explanation:
The derivative of the denominator is 1 so I ignore it.
4 is a constant so I ignore it.
2007-09-25 13:39:08 · answer #1 · answered by UnknownD 6 · 0⤊ 0⤋
You can play around with the factoring:
(4 - s^(1/2)) / (s - 16)
-(s^(1/2) - 4) / (s - 16)
-(s^(1/2) - 4) / (s^(1/2) + 4)(s^(1/2) - 4)
-1 / (s^(1/2) + 4)
The limit of this as s goes to 16 is easy to evaluate now.
2007-09-25 20:40:12 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Appears Simple, Apply L'hopital's Rule. Differentiate numerator and denominator seperately wrt s.
You will get, -(1/(2 sqrt S)).
Apply limit, and the answer is -1/8.
2007-09-25 20:39:23 · answer #3 · answered by Curiousone 2 · 0⤊ 0⤋
It is indeterminate in form 0/0 so use L'Hospital's rule
= lim s--> 16 (-.5/s^.5)/1 = -.5/4 = -.125
2007-09-25 20:43:43 · answer #4 · answered by ironduke8159 7 · 0⤊ 0⤋
(4 - sqrt(s))/(s-16) = (4-sqrt(s))/(sqrt(s)-4)(sqrt(s)+4) =
-1/(sqrt(s)+4) (for s != 16)
so as s approaches 16, f(s) approaches -1/8
2007-09-25 20:40:16 · answer #5 · answered by holdm 7 · 0⤊ 0⤋