# Calculus Answer Confirming Not Sure Im Right Help?

Evaluate the lim

a. lim x--> 64 (cube root x-4/x-64)

(∛x-4)/(x-64) -> 0/0

so then

let cube root x = u

u-4/u^3-64

u-4/u^3-64 = u-4/u-4(u^2+4u+16)

the u-4 cancel each other out leaving

lim x->64 = 1/u^2+4u+16

1/64^2+4(64)=16

oddly i find the number to large am i doing this right?

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1. * 1/64^2+4(64)+16

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2. let ∛x = u
then x = u^3
and as x---> 64, u ---> 4

lim (∛x-4)/(x-64) , as x --->64
= lim (u - 4)/(u^3 - 64) , as u ---> 4
= lim (u-4)((u-4)(u^2 + 4u + 16) , u -->4
= lim 1/(u^2 + 4u + 16), as u --> 4
= 1/(16 + 16 + 16)
= 1/48

When you make your switch from x to u, you also have to make the change in the approach value

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3. Ok thanks for the help,it helped a lot!!!

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4. would the lim approaches 64 is it 1/48 or is it when the lim approaches 4 is 1/48

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5. For the given substitution , the two statements:

lim (∛x-4)/(x-64) , as x --->64
and
lim (u - 4)/(u^3 - 64) , as u ---> 4

are equivalent, so

lim (∛x-4)/(x-64) , as x --->64
= 1/48
and
lim (u - 4)/(u^3 - 64) , as u ---> 4
= 48

They are equivalent questions, we made them that way using the substitution.

here is a little trick with your calculator.
pick a value of x close to the original approach value
e.g. let x = 64.0001
now sub into the original expression
(∛64.0001 - 4)/(64.0001 - 4)
= .000003083/.0001
= .0208332
and 1/48 = .0208333
Now that is not bad!!!!

I used to encourage my students to use this to check their answer.
You could also do this before you work out the question to predict your answer.

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