Who knows how to do limits?

skiah2489

skiah2489

Member
how would you solve this limit:

limit: 3- the square root of(9-x)

x->0 __________________________

x

so in words: the limit of 3 minus the square root of (9-x) all divided by x, as x approaches 0
 
that's easy, just multiply 3 by pi and then get the tangent of that. Square it and divide by 4.
 
works out to 0/0, so do l'hopital's rule, take the derivative of the top and bottom, and apply the limit again.
 
try multiplying the top and bottom by 3 + square root of 9-x ( the conjugate), cancel everything and then plug in the 0
 
im pretty sure the answer is just 0. you plug in the zero, 9-0 =0 therefore the square root of 9=3, 3-3=0 im not 100% sure but thats how i would do it
 
the whole point is that if you plug in the zero for the x on the bottom it becomes zero, so you have to multiply by the conjugate and get the x's to cancel
 
Use L'Hospital's rule, so derive top and bottom, get (1/2)(9-x)^(-1/2) over 1, or simplified to 1/2(9-x)^(1/2). Plug in 0 and get 1/6.
 
you actually multiply by the conjugate

but its -1/6 because 3 minus thesquarerootof(9-x) times 3+ the squarerootof (9-x) equals 9 minus 9 minus x. so the numerator is negative

and the answer is negative instead of positive
 
if she's doing limits she doesn't know what a derivative is (i'm guessing) so she doesn't know how to apply that guys rule...so you multiply by the conjugate
 
equals 9 - (9-x) = x

the numerator is x, the denominator is x(3 + ?(9-x) )

so we have x / ( x ( 3 + ?(9-x) ) )

which simplifies to 1 / ( 3 + ?(9-x) )

apply the limit, the answer is +ve 1/6
 
1)if you plug in 0 to the original equation, you get 0 over 0, that makes it a L'Hospital rule.

2) you forgot that derivative of (9-x)^-.5 equals -.5(9-x)^(-.5) times -1 since you have to chain rule it. and you did the conjugate wrong. (3-(9-x)^(.5))times (3+(9-x)^(.5)) equals 9-(9-x) which equals just x, not -x.
 
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