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

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Posts: 503

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

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

Posts: 7013

that's easy, just multiply 3 by pi and then get the tangent of that. Square it and divide by 4.

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Posts: 18388

exactly

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Posts: 897

fucccckkkk.

how does anyone know how to do that

how does anyone know how to do that

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Posts: 503

oh...

well i somehow got -1/6

i think its right.... am i right?

well i somehow got -1/6

i think its right.... am i right?

Posts: 330

works out to 0/0, so do l'hopital's rule, take the derivative of the top and bottom, and apply the limit again.

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i do!

Posts: 4726

graph it on your calculator. I probably failed my calculus final today so im pretty useless

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Posts: 4726

something like 4/25????

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Posts: 1190

try multiplying the top and bottom by 3 + square root of 9-x ( the conjugate), cancel everything and then plug in the 0

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Posts: 399

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

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Posts: 1190

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

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Posts: 450

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.

Posts: 18388

what?

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Posts: 503

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

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

Posts: 6727

i may have failed math miserable but im pretty sure 9-0 = 9 not 0

Posts: 1236

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

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Posts: 2959

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

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

Posts: 450

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.

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.

Posts: 5647

dont you do *f(x+h)-f(x)*

h

then reduce, then plug in 0 for that equation you reduced

h

then reduce, then plug in 0 for that equation you reduced

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Posts: 5647

that "h" should be below the f(x+h)-f(x) like a fraction

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Posts: 503

yeah

it is 1/6

because the minus sign distributes

okay

thank you everyone

it is 1/6

because the minus sign distributes

okay

thank you everyone

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Posts: 503

r u coming to high north??

Posts: 977

use L'Hospital's rule, that is the simplest way to solve it.

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