Solving "lim x->2 (t^3 - 8)/(t^4 - 16)

[faen]

Homework Statement

lim x->2 (t^3 - 8)/(t^4 - 16)

Homework Equations

The Attempt at a Solution

Well, i just can't find the common factor in the numerator and denomenator. I can split (t^4 - 16) to (t^2 - 2)^4 but i can't find any other factors in the numerator.

Thx for any help :)

 

[emyt]

Homework Statement

lim x->2 (t^3 - 8)/(t^4 - 16)

Homework Equations

The Attempt at a Solution

Well, i just can't find the common factor in the numerator and denomenator. I can split (t^4 - 16) to (t^2 - 2)^4 but i can't find any other factors in the numerator.

Thx for any help :)

multiply by a conjugate or try (t^2+4)(t^2-4)

 

[Fightfish]

There's a rather obvious common factor there. Try factorising 8 and 16.
Alternatively you could use l'Hôpital's rule

 

[faen]

Yeah i tried (t^2+4)(t^2-4), but i can't find the same factor in the numerator. The t^3 term complicates the matter cause what can i multiply with itself to get t^3 without getting a more complicated factor.. In other words, I am unable to express; (t^3 - 8) in any other way.

I tried conjugate but didnt work for me, and we didnt learn l'hospital's yet so not allowed to ues it.

 

[Mark44]

Homework Statement

lim x->2 (t^3 - 8)/(t^4 - 16)

Homework Equations

The Attempt at a Solution

Well, i just can't find the common factor in the numerator and denomenator. I can split (t^4 - 16) to (t^2 - 2)^4 but i can't find any other factors in the numerator.

Thx for any help :)

t⁴ - 16 $\neq$(t² - 2)⁴
Note that the polynomial on the left side is of degree 4, while the one on the right is of degree 8. That should have been a clue that something is wrong.

You should be thinking "difference of squares" and "difference of cubes" for your factoring.

 

[Mark44]

Yeah i tried (t^2+4)(t^2-4), but i can't find the same factor in the numerator. The t^3 term complicates the matter cause what can i multiply with itself to get t^3 without getting a more complicated factor..

t² - 4 can be factored. Also, the difference of cubes can be factored. a³ - b³ = (a - b)(a² + ab + b²).

I tried conjugate but didnt work for me, and we didnt learn l'hospital's yet so not allowed to ues it.

 

[faen]

t⁴ - 16 $\neq$(t² - 2)⁴
Note that the polynomial on the left side is of degree 4, while the one on the right is of degree 8. That should have been a clue that something is wrong.

You should be thinking "difference of squares" and "difference of cubes" for your factoring.

Ah, yeah, i was thinking that it would equal to (t^2 - 4)^2 but the minus sign would be different among the factors.

Anyway i still can't figure it out, can u help me a bit more :p?

 

[faen]

t² - 4 can be factored. Also, the difference of cubes can be factored. a³ - b³ = (a - b)(a² + ab + b²).

now i think i got it.. thanks a lot! :D

 

[faen]

Ok i found that (t-2)(t^2 +2t +4) = t^3 - 8, and now the two t-2 factors in the numerator and denomenator cancels. however I am still stuck with (t^2 - 4) factor which tends to 0 while t tends to 2. So I am stuck again.

If someone could just solve the: lim x->2 (t^3 - 8)/(t^4 - 16) itd be of great help.

 

[Mark44]

Ok i found that (t-2)(t^2 +2t +4) = t^3 - 8, and now the two t-2 factors in the numerator and denomenator cancels. however I am still stuck with (t^2 - 4) factor which tends to 0 while t tends to 2. So I am stuck again.

If someone could just solve the: lim x->2 (t^3 - 8)/(t^4 - 16) itd be of great help.

Then your factoring of t⁴ - 16 is incorrect. Show me how you factored this.

 

[faen]

you are right, ok i finally solved it. Thanks :)