## Babynini one year ago Limits!

1. Babynini

2. Babynini

@freckles :)

3. Babynini

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

4. unimatix

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

5. unimatix

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

6. Babynini

approaching 4 yeah?

7. unimatix

yes number approaching 4

8. Babynini

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

9. Babynini

does 5.33 look right?

10. Babynini

hrm, it's saying that's wrong.

11. unimatix

5.333(repeating) is right believe.

12. unimatix

Maybe try imputing 16/3

13. Babynini

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

14. unimatix

You're welcome :)

15. jim_thompson5910

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

16. jim_thompson5910

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)

17. Babynini

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