Another Infinite Limit - Quite stuck

[Asphyxiated]

Homework Statement

$$\lim_{t \to \infty} \frac {5t+2}{t^{2}-6t+1}$$

Homework Equations

The Attempt at a Solution

Like the title says i am quite stuck, I don't see the first move to make, so if someone could point me in the right direction instead of telling me the answer it would be appreciated.

I tried to factor the bottom trinomial but it wouldn't help out anyway, and fatoring out a t from the top and either a t or t^{2} from the bottom doesn't seem to get me any where. And the answer is suppose to be 0, according to the book.

thanks again!

 

[Tedjn]

You can start in a similar way to your last problem and factor out a t from both top and bottom. Look then at what the numerator and denominator tend to when t approaches infinity.

 

[Asphyxiated]

ok well if you go at it that way it will be:

$$\lim_{t \to \infty} \frac {t(5+2t^{-1})}{t(t-6+t^{-1})}$$

$$\lim_{t \to \infty} \frac {5+2t^{-1}}{t-6+t^{-1}}$$

which is saying that when:

$$H= \infty \;\;\;\;\;and\;\;\;\;\; \epsilon=\frac{1}{H} = infinitesimal$$

it is:

$$\frac {5 + \epsilon}{H-6+\epsilon}$$

which is basically 5 divided by infinity, I don't see why that that equals 0

according to my notes an a constant number divided by an infinite number is going to be an infinitesimal number, is that why the limit is 0? because

$$\frac {5}{H} = \epsilon$$

and epsilon is going to be infinitely close to 0?

 

[Mark44]

The larger H gets, the smaller epsilon gets, so yes, 5/H can be made arbitrarily close to zero.

 

[Asphyxiated]

ok thanks for the help again guys