limits at infinity: (1/3t^5 + 2t^3 - t^2 + 8)

[mikagurl]

Hi all. This may be a crazy question but I am stuck. The problem is:

Evaluate the limit of (1/3t⁵ + 2t³ - t² + 8) as the limit is approaching - infinity.

This is what I have so far:

I factored t⁵ out and I now have [t⁵(1/3 + 2/t² - 1/t³ + 8/t⁵)]. Again the limit is going to - infinity. So I think that all of the terms are approaching - infinity with the exception of 2/t². If we take a really large negative number and square it then would that term go to positive infinity instead? Or does this not matter?

 

[Ishuda]

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Hi all. This may be a crazy question but I am stuck. The problem is to evaluate the limit of (1/3t⁵+2t³-t²+8) as the limit is approaching - infinity. This is what I have so far. I factored t⁵ out and I now have [t⁵(1/3+2/t²-1/t³+8/t⁵)]. Again the limit is going to - infinity. So I think that all of the terms are approaching - infinity with the exception of 2/t². If we take a really large negative number and square it then would that term go to positive infinity instead? Or does this not matter?

If 'things are nice' which generally means individual items go to a limit themselves,
lim (a + b ) = lim a + lim b
and
lim a b = lim a lim b
Applying those to your problem we have, as you started,
\(\displaystyle lim_{t\to-\infty} t^5 (\frac{1}{3} + 2/t^2 - 1/t^3 + 8/t^5) = \frac{1}{3} lim_{t\to-\infty} t^5\)
Since one over a large number is close to zero whether the number is positive or negative. Thus since t⁵ goes to minus infinity as t goes to minus infinity, the limit of the whole expression goes to minus infinity.