Question

Limits!

Answer 1

@freckles :)

Answer 2

So direct substitution didn't work.
Then I tried
((t-4)^4)/((t-4)^3)
Does that work? where would I go from there?

Answer 3

You want to plug in values increasing close to 4 for t.

Answer 4

like 3.9, 4.1
then
3.99, 4.01
etc...
until you see the pattern and there is your answer

Answer 5

approaching 4 yeah?

Answer 6

yes number approaching 4

Answer 7

Ok, let me put it into my calc, just a second :)

Answer 8

does 5.33 look right?

Answer 9

hrm, it's saying that's wrong.

Answer 10

5.333(repeating) is right believe.

Answer 11

Maybe try imputing 16/3

Answer 12

Oh yay! It accepted 16/3
Thanks so much :)

Answer 13

You're welcome :)

Answer 14

t^4 - 256 = (t^2)^2 - (16)^2
t^4 - 256 = (t^2 - 16)(t^2 + 16)
t^4 - 256 = (t - 4)(t+4)(t^2 + 16)


t^3 - 64 = t^3 - 4^3
t^3 - 64 = (t-4)(t^2 + 4t + 16) ... difference of cubes rule


So,

\[\Large \frac{t^4-256}{t^3-64} = \frac{(t - 4)(t+4)(t^2 + 16)}{(t-4)(t^2 + 4t + 16)}\]
\[\Large \frac{t^4-256}{t^3-64} = \frac{\cancel{(t - 4)}(t+4)(t^2 + 16)}{\cancel{(t-4)}(t^2 + 4t + 16)}\]
\[\Large \frac{t^4-256}{t^3-64} = \frac{(t+4)(t^2 + 16)}{t^2 + 4t + 16}\]

Answer 15

after that, you can plug in t = 4 because there are no division by zero errors any more (the t-4 term in the denominator is gone)

Answer 16

ahh that makes sense!! Thanks so much. Yeah, I was trying to figure out how to do it algebraically.