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How do you find #lim sqrt(x+2)-sqrtx# as #x->oo#?

1 Answer

Jan 22, 2017

#lim_(xrarroo)(sqrt(x+2)-sqrtx)# has indeterminate initial form #oo-oo#. See below.

Explanation:

#(sqrt(x+2)-sqrtx) = ((sqrt(x+2)-sqrtx))/1 ((sqrt(x+2)+sqrtx))/((sqrt(x+2)+sqrtx))#

# = (x+2-x)/(sqrt(x+2)+sqrtx)#

# = 2/(sqrt(x+2)+sqrtx)#

As #xrarroo#, the denominator #sqrt(x+2)+sqrtx rarr oo#

Therefore,

#lim_(xrarroo) (sqrt(x+2)-sqrtx) =lim_(xrarroo) 2/(sqrt(x+2)+sqrtx)#

# = 0#

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