Question

How would I evaluate this limit

Answer 1

divide x^4 top and bottom

Answer 2

Alright I didn't think of that haha just a usual habit to multiply by conjuage..

Answer 3

i tried, its not working :|

Answer 4

uhh.its not working

Answer 5

dividing x^4 is a fail it seems

Answer 6

multiplying conjugate is a mess, but seems it'll work, do it one by one...

Answer 7

I \[\frac{ --- }{ x-\sqrt{x^{2}+\frac{x}{2}+1} } * \frac{ ---- }{ x+\sqrt{x^{2}+5x+2} }\]get stuck at the part whre

Answer 8

ok, so numerator first,
\((a+b)(a-b)=a^2-b^2\)
\(x^2-(x^2+5x+2)\)

Answer 9

I got that

Answer 10

I have a few of steps to the way I did it.

Answer 11

keep the denominator (x+sqrt...) as it is
and try conjugate for denom. now...

Answer 12

I will let you guys do whichever way you are trying. I will be back.

Answer 13

what I have it like this and now I don't know what to do

Answer 14

I did two things of rationalizing.

Answer 15

\[\frac{x-\sqrt{x^2+5x+2}}{x-\sqrt{x^2+\frac{x}{2}+1}} \cdot \frac{x+\sqrt{x^2+5x+2}}{x+\sqrt{x^2+\frac{x}{2}+1}} \cdot \frac{x+\sqrt{x^2+\frac{x}{2}+1}}{x+\sqrt{x^2+5x+2}}\]

Answer 16

\[\frac{x^2-(x^2+5x+2)}{x^2-(x^2+\frac{x}{2}+1)} \cdot \frac{x+\sqrt{x^2+\frac{x}{2}+1}}{x+\sqrt{x^2+5x+2}}\]
Now find the limt of the first part and find the limit of the second part
the limit of the whole thing will be the product of those limits

Answer 17

brilliant ! wolfram dumped 10 pages for this same solution !!

Answer 18

10 pages?

Answer 19

I'm not making an account so I won't look.

Answer 20

What! I don't have space for 10 pages of solutions

Answer 21

I have to give it in the limited space and I have like 5 min per question.

Answer 22

No. This way I'm telling you to evaluate won't let to 10 pages.

Answer 23

lead*

Answer 24

He was just saying wikepedia did it a 10 page way. I'm totally not doing a 10 page way. Those most this problem should take up on your page is 4 lines depending how big or small you write.

Answer 25

Have you tried doing what I said?

Answer 26

yea on it

Answer 27

so I got this

Answer 28

That's good.
I hope you look at it in parts like I said:
\[\lim_{x \rightarrow \infty}\frac{x^2-(x^2+5x+2)}{x^2-(x^2+\frac{x}{2}+1)}\]
and
\[\lim_{x \rightarrow \infty}\frac{x+\sqrt{x^2+\frac{x}{2}+1}}{x+\sqrt{x^2+5x+2}}\]

First part should be pretty easy...

Second part...We divide by sqrt(x^2) which equals x since x>0

Answer 29

I was typing part of this before you sent that. lol.

Answer 30

It was good...but there are even harder questions on the exam T_T... I will keep you guys updated. thanks guys ! @myininaya @hartnn @ganeshie8

Answer 31

That was a good question. Unlike the one you showed before, I haven't this one.
I have seen similar problems where we had to do a 2 stepper of rationalizing.

Answer 32

Guys... I just turned the page and now its even worse....

Answer 33

lol/ That just means more fun.

Answer 34

new post im closing this

Answer 35

@hartnn Hey!

Coming back to this limit. How did we know that you can't get the correct limit using the divide top and bottom by \[\frac{1}{x^{4}}\]

Answer 36

like when I was doing the math for that question I got 1 which was the wrong answer. But, In a test how would I know that I am doing it wrong?

Answer 37

we will not know before hand what will work....we just need to try out different things...

Answer 38

but my answer for 1/x^4 = 0 looked mathematically correct. Why was it wrong?

Answer 39

is there a mathematical thing we can determine... can we just plug it in and like see... but it was going to be 0/0 anyhow or infinity/infinity ..

Answer 40

@hartnn why wouldn't the answer of 1 be correct... It is only a factor of 10 lower than the real solution

Answer 41

We didn't want the bottom to approach 0 or infinity.
We wanted the bottom to approach some constant number.
To get some constant number on bottom we divide by 1/(sqrt{x^2}
or 1/x in this case since x>0.

Answer 42

We couldn't go ahead and use this from the beginning because we would have gotten that the top approached 1-sqrt(1)
and the bottom approached 1-sqrt(1)
This would be 0/0

So that is why I decided to multiply both top and bottom by top's conjugate and bottom's conjugate.
Then I tried again to divide top and bottom by 1/(sqrt(x^2)) which equals 1/x since x>0.