Limits,
http://screencast.com/t/MeamUq9EgG8h
I heard there is a rule about "powers" when it comes to limits like this in order to solve them? Can you tell me the rule please?

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- Christos

- schrodinger

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- UnkleRhaukus

Is this your question ?\[\large\lim_{x\to+\infty}\frac{3x+1}{2x-5}\]

- Christos

yes

- zepdrix

Yes there is a nice little shortcut, which is what you must be referring to.
I'm trying to think of how to explain it.
Since it's an infinite limit, we only need to worry about the `leading` term (highest power on top and bottom).
Compare them, if they're equal degree, the limit will approach the ratio of their coefficients. In this case, 3/2.

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## More answers

- UnkleRhaukus

when \(x\to+\infty\), the ratio will ok like \(\frac{3x}{2x}\)

- Christos

you mean like 3/2 ??

- Christos

and when x->-inf -3/2 ?

- zepdrix

Both x's are approaching negative infinity, so you can think of the negatives cancelling each other out, giving us the same 3/2 as with before :)

- Christos

What about higher power at the denominator or the numerator separately

- Christos

@zepdrix ?

- zepdrix

It might help if you see an example and just think about it logically.
\[\large \lim_{x \rightarrow \infty}\frac{\color{green}{x^2-2x+3}}{\color{royalblue}{x-7}}\]
The green part represents a parabola,
while the blue part a line.
Which one will grow towards infinity faster?
A straight line or a parabola?

- Christos

1 ?

- zepdrix

what? XD lol

- Christos

uhm I think its the whole numerator? I am not sure :D

- zepdrix

noooo it's not :O
See how the `power` on x is larger in the top?
We can't apply the same rule we used before, since the highest powers of x `do not match`.

- zepdrix

|dw:1369665937413:dw|Which one of these functions is growing upwards faster? :O

- zepdrix

The curve or the straight line? :D

- Christos

numerator = denominator?

- Christos

Eeeh i mean that high one not the straight one the other one grows faster

- zepdrix

Yes good good, the curvy one grows faster.

- zepdrix

\[\large \lim_{x \rightarrow \infty}\frac{\color{green}{x^2-2x+3}}{\color{royalblue}{x-7}}\]The `green` part is THAT type of function, the curvy one.
So we can think of it this way, as the numerator and denominator approach infinity, the numerator is "winning", it's getting there faster.
So the limit will approach infinity.

- zepdrix

That too confusing? :3 stick with the math rules?

- Christos

same applies for x->-inf ? and what about the denominator having the highest power?

- zepdrix

Let's look at the reverse a moment.
\[\large \lim_{x \rightarrow \infty}\frac{\color{royalblue}{x-7}}{\color{green}{x^2-2x+3}}\]
Now the BOTTOM is a larger degree than the top.
So the bottom is "winning", approaching infinity faster.
So our limit will approach \(\large \dfrac{1}{\infty}\)
which is the same as saying, it's approaching zero.
The denominator is getting bigger and bigger faster than the top, so the fraction becomes smaller and smaller.

- Christos

thanks

- zepdrix

Negative infinity is a little trickier, since even powers will mess things up sometimes... I'm not sure how to explain that part properly :3

- Christos

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Congrats!

- UnkleRhaukus

Well done @Christos !
\(\LARGE\mathcal{CONGRATULATIONS!!}\)

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