Lim (2+√x)/(2-√x)
x->∞
PLEASE HELP! Try not to use the l' Hospital's rule.
Multiply by (2-√x) / (2-√x)
(2+√x) (2-√x) / (2-√x) (2-√x) ----- (1)
The numerator is of form (a+b)(a-b) = a^2-b^2
a=2
b= √x
a^2-b^2 = 4-x
(1) becomes:
(4-x) / (2-√x)^2
= x(4/x-1) / x(2/√x -1)^2
= (4/x-1) / (2/√x -1)^2 --- (2)
lim x-->∞ 4/x=0
lim x-->∞ 2/√x =0
(2) becomes:
-1/(-1)^2 = -1
The limit is -1
-1
divide every term by root x so that terms 2/root x converge to zero, while the root over root is 1 so that 1/-1 = -1
hi :)
without de l'hopital:
http://www.xlogx.com/en/limit-of-a-function/26b88d...
Hope this helps! Bye!
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Verified answer
Multiply by (2-√x) / (2-√x)
(2+√x) (2-√x) / (2-√x) (2-√x) ----- (1)
The numerator is of form (a+b)(a-b) = a^2-b^2
a=2
b= √x
a^2-b^2 = 4-x
(1) becomes:
(4-x) / (2-√x)^2
= x(4/x-1) / x(2/√x -1)^2
= (4/x-1) / (2/√x -1)^2 --- (2)
lim x-->∞ 4/x=0
lim x-->∞ 2/√x =0
(2) becomes:
-1/(-1)^2 = -1
The limit is -1
-1
divide every term by root x so that terms 2/root x converge to zero, while the root over root is 1 so that 1/-1 = -1
hi :)
without de l'hopital:
http://www.xlogx.com/en/limit-of-a-function/26b88d...
Hope this helps! Bye!