How do you evaluate the limit #sqrt(x-2)-sqrtx# as x approaches #oo#?

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Answer 1 · 2016-10-04T13:59:49

0

#lim_(x to oo) sqrt(x-2)-sqrtx#

#= lim_(x to oo) sqrt x sqrt(1-2/x)-sqrtx#

#= lim_(x to oo) sqrt x ( (1-2/x)^(1/2) - 1 )#

expanding by Binomial Expansion

#= lim_(x to oo) sqrt x ( 1-1/2 2/x +O(1/x^2 )- 1 )#

#= lim_(x to oo) -1 /(sqrt x) + O(1/x^(3/2) ) = 0#

Answer 2 · 2016-10-05T21:20:20

#0#

#lim_(xrarroo)sqrt(x-2)-sqrtx#

We can rewrite this square root using its conjugate:

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

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

We can try to take a factor from the denominator:

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

#=lim_(xrarroo)(-2)/(sqrtxsqrt(1-2/x)+sqrtx)#

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

Notice that as #xrarroo#, we see that #2/x# goes to #0#. So, we can "evaluate" this limit:

#=lim_(xrarroo)(-2)/(sqrtoo(sqrt(1-0)+1))#

#=lim_(xrarroo)-1/sqrtoo#

#sqrtoo# is still infinite, so just like how #2/x# approached #0#, this will also tend towards #0#.

#=0#