Question

How do you find the limit when x is approaching 0, (x^2+3)/x^4? @welshfella

Answer 1

@Phi

Answer 2

@zepdrix anyone out there

Answer 3

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

Answer 4

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

Answer 5

so ummmmmm :)

Answer 6

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

Answer 7

@zepdrix yes

Answer 8

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

Answer 9

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

Answer 10

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.

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

@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?

Answer 16

@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.

Answer 17

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\[\large\rm \lim_{x\to0}\frac{x^2+3}{x^4}\left(\frac{1/x^4}{1/x^4}\right)=\lim_{x\to0}\frac{\frac{1}{x^2}+\frac{3}{x^4}}{1}=\lim_{x\to0}\frac{1}{x^2}+\lim_{x\to0}\frac{3}{x^4}\]

Answer 18

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

Answer 19

That seems wayyy tedious though :d
Intuitionnnnnnn instead!

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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.

Answer 24

Blah, math is hard >.<