lim t-->1 t^3/ (t^2-1)^2

I just really want to know if you have to use algebra to factor out the denominator or do i just plug in the values i.e. 0.9 from the left and 1.1 for the right

NEED FOR CLASS AT 10 THIS MORNING!

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- May 23rd 2011, 05:21 AMpurplec16infinite limits
lim t-->1 t^3/ (t^2-1)^2

I just really want to know if you have to use algebra to factor out the denominator or do i just plug in the values i.e. 0.9 from the left and 1.1 for the right

NEED FOR CLASS AT 10 THIS MORNING! - May 23rd 2011, 05:24 AMAckbeet
The numerator goes to 1, and the denominator goes to zero. Since the denominator is always positive, the limit is positive infinity. Done.

- May 23rd 2011, 05:29 AMpurplec16
I know the answer i have to do the working using first principle and I want to know how to work it for when the lim t-->1 is coming from the left and for example i use the value 0.9 do i put it in as

(0.9)^3/ ((0.9)^2-1)^2 or (0.9)^3/(0.9-1)(0.9+1) - May 23rd 2011, 05:43 AMDeveno
"plugging in values" doesn't evaluate the limit, in some unusual cases, a function can fluctuate quite wildly near the point the limit is evaluated at.

in your case, what you would need to show that no matter how big a number N you choose, you can pick a value of t close enough to 1 so that

t^3/(t^2 -1)^2 > N. the value that will work, will depend on the number N. - May 23rd 2011, 03:20 PMpurplec16
Thank You so much I wasn't able to see your reply before class but I plugged in the answer and got the value