Expert: Paul Klarreich - 4/14/2007

Question
L'Hospital's rule does not help with the following limits. find them some other way a) LIM as x approaches infinity (x+5)^1/2/(x)^1/2 +5    b) 2x/x +7(x)^1/2

Answer
Questioner:   nyawas
Category:  Calculus
 
Subject:  calculus
Question:  L'Hospital's rule does not help with the following limits. find them some other way a) LIM as x approaches infinity (x+5)^1/2/(x)^1/2 +5    b) 2x/x +7(x)^1/2
.....................................
Hi, Nyawas,

Whenever someone says ".... doesn't help" I usually answer, Oh, yeah?  Says you!

However, I think you might be correct about these.  Here's a general clue:

Suppose you have something like:
       Polynomial in x, of degree n
lim     -------------------------------
x->inf  Another Poly in x, also degree n

Scheme:  Divide top and bottom by x^n, so you get things like:

17 + 23/x + ...  +7/x^n
-----------------------------
similar stuff on the bottom

Now things that have x's in the denominator, like 23/x, become zero at infinity.  So only constant terms survive.

................................................
Now to your examples.

I AM GUESSING ABOUT YOUR EXAMPLES.  YOUR PARENTHESIZING IS NOT SO CLEAR.  AND REMEMBER THIS BASIC:

WARNING: THIS DISCUSSION MAY CONTAIN FRACTIONS AND OTHER MATERIAL DIFFICULT TO VIEW ON CERTAIN COMPUTING SYSTEMS.  VIEW IT IN A FIXED-SIZE FONT, SUCH AS COURIER.


        sqrt(x + 5)
lim      -------------
x->inf   sqrt(x) + 5

Divide top and bottom by  sqrt(x):

       sqrt(x + 5)/sqrt(x)
lim     ---------------------
x->inf  (sqrt(x) + 5)/sqrt(x)

       sqrt(1 + 5/x)
lim     --------------
x->inf  1 + 5/sqrt(x)

 sqrt(1 + 0)
= ----------- = 1
 1 + 0

................................

             2x
lim      -------------
x->inf   x + 7 sqrt(x)

Divide top and bottom by x:

             2
lim      -------------
x->inf   1 + 7/sqrt(x)
   2
= ----- = 2
 1 + 0