faen said: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 [itex]\neq[/itex](t2 - 2)4faen said: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 :)
t2 - 4 can be factored. Also, the difference of cubes can be factored. a3 - b3 = (a - b)(a2 + ab + b2).faen said: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..
faen said:I tried conjugate but didnt work for me, and we didnt learn L'Hopitals yet so not allowed to ues it.
Mark44 said:t4 - 16 [itex]\neq[/itex](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 said:t2 - 4 can be factored. Also, the difference of cubes can be factored. a3 - b3 = (a - b)(a2 + ab + b2).
faen said: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.
The given limit is 2.
To solve this limit, you can use algebraic manipulation and the properties of limits. First, factor the numerator and denominator to get (t-2)(t^2+2t+4)/(t-2)(t^3+2t^2+4t+8). Then, simplify the expression by cancelling out the common factor of (t-2). This leaves you with (t^2+2t+4)/(t^3+2t^2+4t+8). Finally, plug in the given limit of 2 to get the answer of 1/4.
Limits are important in calculus because they allow us to describe the behavior of a function as it approaches a certain point. In this case, the limit is used to find the value of the function at the specific point of x=2.
Yes, you can use a graph to help you visualize the behavior of the function as it approaches the given limit. However, algebraic manipulation and the properties of limits are still necessary to find the exact value of the limit.
Yes, there are special cases for solving this limit. One special case is when the denominator is equal to 0. In this case, the limit does not exist. Another special case is when the numerator and denominator are both equal to 0. In this case, you may need to use L'Hopital's rule to solve the limit.