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- anonymous

- anonymous

@zepdrix anyone out there

- anonymous

You could graph it and look, that's always a good place to start

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## More answers

- anonymous

@RunawayGalaxy I honestly don't even know how to graph that.

- zepdrix

so ummmmmm :)

- zepdrix

\[\large\rm \lim_{x\to0}\frac{x^2+3}{x^4}\]Simply plugging in 0 gives you an indeterminate form of \(\large\rm \frac{3}{0}\) ya?

- anonymous

@zepdrix yes

- zepdrix

Do you remember anything about this limit?\[\large\rm \lim_{x\to0}\frac{1}{x}\]Because our problem is behaving the same way.

- anonymous

I do not remember much about that particular limit. Would I replace x with 0? 1/0 is undetermined?

- zepdrix

Well let's plug in a value that's very close to 0, that should give us an idea of what's going on.
How about 1/99999, that's pretty close to 0 right? :)\[\large\rm \lim_{x\to0}\frac{1}{\color{orangered}{x}}\approx\frac{1}{\color{orangered}{\frac{1}{99999}}}=1\cdot\frac{99999}{1}=99999\]Do you understand what I did here?
I plugged in a really really small fraction, a value close to zero.
And applying an algebra rule, the result is giving us a really big value.

- zepdrix

So as x is getting closer to 0, 1/x is growing infinitely large.

- anonymous

@zepdrix Does L'hopital's rule not apply in this case? Is 3/0 not indeterminate form?

- zepdrix

It's not the indeterminate form that we need for L'Hop unfortunately :(
Need 0/0 or infty/infty

- anonymous

I remembered that there were a bunch, but couldn't figure if that was one of them

- zepdrix

@raesal what do you think? :d
do you understand why 1/x is blowing up large and in charge like that? :D
make sense of the example?

- anonymous

@zepdrix I understand the example, I just don't understand how that example is going to help me solve this problem. I don't understand how 1/x "blowing up large and in charge" will help me solve this problem.

- zepdrix

Well it's blowing up like, is partly explained by the fact that it's approaching the indeterminate form 1/0.
Notice that with our problem, we're approaching the indeterminate form 3/0.
So we will end up with the same result.
I'm not sure of the exact algebra steps.. umm
It's easier to think about it intuitively.
The bottom is a 4th power, so it's doing everything "faster" than the top.
It's getting to zero faster than the numerator, so it blows up just like 1/x.
Oh oh maybe we could do this: Multiply top and bottom by 1/x^4\[\large\rm \lim_{x\to0}\frac{x^2+3}{x^4}\left(\frac{1/x^4}{1/x^4}\right)=\lim_{x\to0}\frac{\frac{1}{x^2}+\frac{3}{x^4}}{1}=\lim_{x\to0}\frac{1}{x^2}+\lim_{x\to0}\frac{3}{x^4}\]

- zepdrix

Then we can relate it back to that 1/x limit in each of these limits,\[\large\rm =\left(\lim_{x\to0}\frac{1}{x}\right)^2+3\left(\lim_{x\to0}\frac{1}{x}\right)^4\]

- zepdrix

That seems wayyy tedious though :d
Intuitionnnnnnn instead!

- zepdrix

That multiplication step was kinda sloppy, i shoulda just separated the fractions, x^2/x^4 and 3/x^4, same result.

- zepdrix

I dunno, it's a lot to take in >.<
Lemme know what you're thinkin

- anonymous

I'm confused. I'm so sorry >.<

- zepdrix

When you have a fraction:\[\large\rm \lim_{x\to 0}\frac{stuff}{other~stuff}\]
If the stuff in the numerator is going towards 0 faster, then the fraction overall is going towards zero. It's approaching this form: \(\large\rm \frac{0}{number}\) which is just zero.
When the bottom is going towards 0 faster, then the whole thing is approaching infinity.
It's approaching this form: \(\large\rm \frac{number}{0}\)
This second case is what is happening with our problem,
the bottom is getting to zero much much faster.

- zepdrix

Blah, math is hard >.<

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