I need to find out for what values of A and B is the statement correct.
lim {under square root (Ax+B)} -3 divided by x = 1
x apporaches 0
Only the Ax + B is under the square root -3 is not under the square root. It's divided by only x.
2007-04-14 06:28:04 · 4 answers · asked by glance 3 in Science & Mathematics ➔ Mathematics
y = [sqrt(Ax+B) -3]/x
as x->0, y-->oo unless y also -->0
so sqrt(Ax+B)-->3 so B=9
lim [sqrt(Ax+9)-3]/x
= lim[ [(1/2)A/sqrt(Ax+9)] / 1 ] ...l'hopital
= (1/2)A/3 = 1
so A = 6
2007-04-14 06:39:28 · answer #1 · answered by hustolemyname 6 · 0⤊ 0⤋
1) If B is not 9 the limit of the numerator would be sqrtB-3 not zero and the limit would be infinity
so B=9
we have now lim [sqrt(Ax+9)-3]/x
If we multiply numerator and denomi
nator
by sqrt(Ax+9)+3 we get lim [(Ax+9)-9]/[x *(sqrt(Ax+9)+3] = A/6 = 1 so A = 6
2007-04-14 13:41:37 · answer #2 · answered by santmann2002 7 · 0⤊ 0⤋
The question stated is a bit weird. Assuming the whole thing is over x, as the others have, the expression tends to A/6 as x->0.
What's the x=1? If x=1 it isn't tending anywhere but 1.
2007-04-14 14:05:08 · answer #3 · answered by anotherbsdparent 5 · 0⤊ 0⤋
? what is under square root?
do you mean
(sqroot (Ax +b) -3) / x
x =1 to x=0
most of the time they are asking for limits
example limit a limit b
there a 1000 answers as question is asked
2007-04-14 13:44:34 · answer #4 · answered by zzjoev 2 · 0⤊ 0⤋