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?
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
* 1/64^2+4(64)+16
Ok thanks for the help,it helped a lot!!!
would the lim approaches 64 is it 1/48 or is it when the lim approaches 4 is 1/48
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.
To evaluate the given limit, you correctly started by substituting `u = cubic root of x`. Then, you simplified the expression to `(u - 4) / (u^3 - 64)`. However, at this point, instead of trying to cancel out `(u - 4)`, you need to analyze the expression further.
First, factor the denominator `u^3 - 64` using the difference of cubes formula: `a^3 - b^3 = (a - b)(a^2 + ab + b^2)`.
So, we have `u^3 - 64 = (u - 4)(u^2 + 4u + 16)`.
Now, we can cancel out `(u - 4)` in the numerator and denominator:
`(u - 4) / (u^3 - 64) = 1 / (u^2 + 4u + 16)`
Now, substitute `u = cube root of x` back into the expression:
`= 1 / ((cube root of x)^2 + 4(cube root of x) + 16)`
`= 1 / (x^(2/3) + 4x^(1/3) + 16)`
Since the limit is as x approaches 64, substitute `x = 64` into the expression:
`= 1 / (64^(2/3) + 4(64)^(1/3) + 16)`
`= 1 / (4 + 16 + 16)`
`= 1 / 36`
Therefore, the limit as x approaches 64 is `1/36`.
Regarding your concern about the number appearing too large, it seems you made a mistake when calculating the value of `64^2 + 4(64)`. Please double-check your arithmetic to ensure accuracy.