Question

limit

Answer 1

\[\lim_{x \rightarrow \infty} \sqrt{x^2+5x-2}-\sqrt{4x^2-3x+7}+\sqrt{x^2+7x+5}\]

Answer 2

what should i do first ?

Answer 3

I guess you need to do multiplication by conjugate

Answer 4

I'm not completely sure.

Answer 5

My intuition tells me all that really matters here is the \(\sqrt{x^2}-\sqrt{4x^2}+\sqrt{x^2}\).

Answer 6

But when I think about it, if those terms cancel out, then the secondary terms become important.

Answer 7

I think there's more to it than that. The answer is a fraction...

Answer 8

actually i can if only 2 root2, but there are 3 roots now....

Answer 9

It's possible factoring could lead to some insite as well.

Answer 10

That second term will be tricky

Answer 11

I lost my work.

Answer 12

not that I would do much anyway

Answer 13

Okay, seriously I think using conjugate here is the answer.

Answer 14

take out the 4 in the second term ? @SithsAndGiggles

Answer 15

@tanjung my point was that the discriminant of the second quadratic is negative.

Answer 16

I'm curious to see if there's some hidden pattern among these coefficients...

Answer 17

\[
(\sqrt{x^2+5x-2}+\sqrt{x^2+7x-5})-\sqrt{4x^2-3x+7}
\]Conjugate is: \[
(\sqrt{x^2+5x-2}+\sqrt{x^2+7x-5})+\sqrt{4x^2-3x+7}
\]

Answer 18

how about this :
\[\sqrt{x^2+...} - \sqrt{4x+...} + \sqrt{x^2+...} = \sqrt{x^2+..}-\sqrt{x^2+...}+\sqrt{x^2+...}-\sqrt{x^2+...}\]

Answer 19

can i group them then multiplr by conyugate :
[lim sqrt(x^2 + ...)-limsqrt(x^2-....)] + [lim sqrt(x^2 + ...)-limsqrt(x^2-....)]

Answer 20

Hmm, completing the square seems to be a step in the right direction. You have
\[x^2+5x-2=\left(x+\frac{5}{2}\right)^2+\cdots\]
observe that
\[\sqrt{x^2+5x-2}-\left(x+\frac{5}{2}\right)+\frac{5}{2}\to\frac{5}{2}\]
as \(x\to\infty\). Try this with the other terms.

Answer 21

Actually at this point you might try the conjugate approach with the remaining two terms.

Answer 22

yeah, if i use the conyugate like i did above, it will be
limx->~ sqrt(x^2+5x-2)-sqrt(4x^2-3x+7)+sqrt(x^2+7x+5)
= limx->~ sqrt(x^2+5x-2)-2sqrt(x^2-3/4x+7/4)+sqrt(x^2+7x+5)
= [limx->~ sqrt(x^2+5x-2)-sqrt(x^2-3/4x+7/4)] +
[limx->~ sqrt(x^2+7x+5)-sqrt(x^2-3/4x+7/4)]
obvious i can do it one by one above, but is this allowed in property of limit ?

Answer 23

That way turns out to be much simpler... And yes, you're not breaking any rules :)

Answer 24

You can also try using the binomial theorem and see which terms approach 0.

Answer 25

ok ok ... btw i get 54/8 or 27/4

Answer 26

Yes that's correct