How Do You Solve These Challenging Calculus Limit Problems?

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few calc limit questions! need help!

Homework Statement



limit x-> -1 of cubed rt (3x-5/25x-2)
not sure how to go about this at all

i think the answer is 2/3 but i can't work it up


lim x->0 √(3x+2) - √2
x


r(x)= |3x| A) lim x->0 of r(x) question 2 B) r(0)
x





The Attempt at a Solution

 
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these weren't formatted right. the middle one the x is underneath the top term as well as for the absolute (|3x|)/x

also, i tried to multiply the middle one by its conjugate √(3x+2) - √2

i wind up getting (3x)/(x (√(3x+2) - √2) ) i think that's the right track but I am not sure
 
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hi venom2121! welcome to pf! :smile:

(btw, never reply to your own first post :redface:, use the EDIT button instead … then you'll stay on the No-replies list! :wink:)
venom2121 said:
limit x-> -1 of cubed rt (3x-5/25x-2)
not sure how to go about this at all

i think the answer is 2/3 but i can't work it up

but it isn't 0/0, so what's the difficulty? :confused:

just put x = -1 ! :rolleyes:
lim x->0 (√(3x+2) - √2)/x

use a binomial expansion (and √(3x+2) = √x√(3 + 2/x)) :wink:
r(x)= (|3x|)/x

try drawing a graph :wink:
 


the cubed rt problem the answer is supposed to be 2/3 DERP ok had a brain fart haha got that now..

im still confused about the binomial expansion. and as far as the absolute value one I am not sure. i see that the y values max out at -3 and 3 and as it goes to zero the y values stay at 3 until zero. so is that the limit? would that mean that r(0)=0? also how would you figure this out without graphing?
 

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venom2121 said:
im still confused about the binomial expansion.

can't you do the binomial expansion of (1 + x)1/2 ?
and as far as the absolute value one I am not sure. i see that the y values max out at -3 and 3 and as it goes to zero the y values stay at 3 until zero. so is that the limit? would that mean that r(0)=0?

not following you :confused:
also how would you figure this out without graphing?

i wouldn't!
 


venom2121 said:

Homework Statement



limit x-> -1 of cubed rt (3x-5/25x-2)
not sure how to go about this at all

i think the answer is 2/3 but i can't work it up


lim x->0 √(3x+2) - √2
x


r(x)= |3x| A) lim x->0 of r(x) question 2 B) r(0)
x





The Attempt at a Solution


Your "cube-root" problem, as written, is
\lim_{x \rightarrow -1} \sqrt[3]{\left( 3x - \frac{5}{25x} - 2\right)} = -\frac{\sqrt[3]{600}}{5},
but perhaps you meant
\lim_{x \rightarrow -1} \sqrt[3]{\left( \frac{3x-5}{25x - 2}\right)}.
In that case you should have used brackets, and written cube rt ((3x-5)/(25x-2)).

RGV
 
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