Question

Heya! Double checking my answer since I think I am doing something wrong. Taking the positive infinite limit of (x^2)/sqr((x^4)+1). I divide by x^4 and eventually get 0/0 or DNE. When I check on the calulator however, it seems to be approaching 1. What am I doing wrong? :)

Answer 1

The problem

Answer 2

have you covered l'hopital's rule?

Answer 3

I read about it online, but not explicitly, no

Answer 4

We aren't "supposed" to know how to take derivativesin this section

Answer 5

thats why I asked :P
\[\lim_{x \rightarrow \infty} \frac{ x^2 }{ \sqrt{x+1} }\]

Answer 6

oops to the power of 4 inside let me rewrite that

Answer 7

\[\lim_{x \rightarrow \infty}\frac{ x^2 }{ \sqrt{x^4+1} }\]

Answer 8

lets think about this

Answer 9

as we approach infinity do we care about the +1?

Answer 10

nope

Answer 11

so and this is not the way to write it out in your book but we can think of it as just the square root of x^4 on the bottom

Answer 12

what is the square root of x^4?

Answer 13

infinity in this case

Answer 14

oh nvm

Answer 15

in general not in this case

Answer 16

is it (x^4)^(1/2)= x^(2) ?

Answer 17

yes

Answer 18

can I do that? I thought because of the constant I can't apply the exponent to simplify

Answer 19

technically we can't do it that way but I am showing you how I think of it. Now to do it more formally we divide both the numerator and the denominator by x^2 not x^4

Answer 20

aha!

Answer 21

remember that intially the divide by x^2 is outside of the square root so when you bring it in it changes to x^4

Answer 22

ok. So a general question then for infinite limits. When we have a square root in the denominator and I am looking for the highest exponent to divide by, I should take sqrt(x^n) ?

Answer 23

I think you should try to think of this in an intuitive kind of way and think, hey, x^2 is only slightly smaller than sqrt(x^4+1) since if that +1 wasn't in there, it'd just be x^2 on bottom too! So as x gets larger and larger, that little 1 is going to be less noticeable until eventually, at infinity, the two values become pretty much the same. And anything divided by itself is just 1.

Answer 24

I would lean more towards the thinking I walked you through and @Kainui just described again. It will help you understand it more than just a rinse and repeat procedure

Answer 25

hmm. I can kind of visualise this in my head. but

Answer 26

I'm still not exactly clear on where the x^2 divisor comes from

Answer 27

whether it's a function of the particular ratio of this function or what

Answer 28

like I said I took x^4 before, but that's obviously wrong

Answer 29

well remember when we said what the square root of x^4 is?

Answer 30

yeah

Answer 31

so what I said was right, but I should also visualize

Answer 32

thats where it came from looking at the new limit of
\[\lim_{x \rightarrow \infty} \frac{ x^2 }{ x^2 }\]

Answer 33

aha I see

Answer 34

where I just subbed in what we said we cared about in the denominator

Answer 35

its kinda weird how the way everyone thinks of it...is not the method you put down on paper lol

Answer 36

but in terms of the steps I show..

Answer 37

lol

Answer 38

so here's how I see it now

Answer 39

first off, drop the 1

Answer 40

then take sqrt(x^4) to get x^2

Answer 41

x^2 / x^2 over x^2 / x^2

Answer 42

which makes 1/1 =1

Answer 43

am I seeing this correctly?

Answer 44

yup thats how I think of it

Answer 45

sweetness

Answer 46

Alright well don't forget to close the question!

Answer 47

yep, thanks