/sci/ - help me with Calc please

Name: Anonymous 2010-08-01 22:32

hey guys, I keep trying to solve this Calc problem.

lim sqrt(x^2+5x)-x
x->[infinity]

I know that it eventually works out to being 5/2. But the some of the steps are a little beyond me. Can anybody help?

Name: Anonymous 2010-08-02 0:55

Rationalize the numerator: Multiply by \frac{\sqrt{x^2+5x}+x}{\sqrt{x^2+5x}+x}.

Name: Anonymous 2010-08-02 16:57

>>1
I would suggest that you invert the cosign.
Also, you'll get a better crop yield if you de-lineate the tangents, and then proceed with mulching.

Name: Anonymous 2010-08-03 6:33

>>3
>cosign
>cosine

Name: Anonymous 2010-08-06 16:55

>>3
back to /farm/, please.

Name: Anonymous 2010-08-07 0:37

>>5
/farm/ is not a board.

Name: Anonymous 2010-08-07 8:35

>>6
http://dis.4chan.org/farm/

Name: Anonymous 2010-08-07 14:18

>>7
404 - Not Found

Name: >>6,8 2010-08-07 14:19

Addendum: Though, /5/ was for a short time.

Name: Anonymous 2010-08-07 21:36

the limit of a power is the power of the limit.

lim[f(x)]^n = [lim f(x)]^n

Name: Anonymous 2010-08-10 5:17

this is easily done by completing the square:

sqrt(x^2+5x) - x = sqrt( (x+(5/2))^2 - (25/4) ) - x

with x->infty, the term under the square root has constant difference to (x+(5/2))^2, therefore the square root is asymptotic to (x+(5/2)). All in all we get:

lim x+5/2 - x = 5/2
x->infty

Name: Anonymous 2010-08-10 10:09

DON'T HELP HIM!!!

Name: Anonymous 2010-08-11 3:34

>>12
That would be rude.

Name: Anonymous 2010-08-12 3:54

>>13
Yes. Helping him would be rude.

help me with Calc please

1 Name: Anonymous 2010-08-01 22:32

2 Name: Anonymous 2010-08-02 0:55

3 Name: Anonymous 2010-08-02 16:57

4 Name: Anonymous 2010-08-03 6:33

5 Name: Anonymous 2010-08-06 16:55

6 Name: Anonymous 2010-08-07 0:37

7 Name: Anonymous 2010-08-07 8:35

8 Name: Anonymous 2010-08-07 14:18

9 Name: >>6,8 2010-08-07 14:19

10 Name: Anonymous 2010-08-07 21:36

11 Name: Anonymous 2010-08-10 5:17

12 Name: Anonymous 2010-08-10 10:09

13 Name: Anonymous 2010-08-11 3:34

14 Name: Anonymous 2010-08-12 3:54