How do you find the limit of #(1/sqrtx-1/sqrt2)/(x-2)# as #x->2#?

1 Answer
Jim H
Oct 29, 2016

Begin by rewriting it without the difference on fractions in the numerator of a fraction. Then see where that leads.

Explanation:

#(1/sqrtx-1/sqrt2)/(x-2) = ((1/sqrtx-1/sqrt2))/((x-2)) * sqrt(2x)/sqrt(2x)#

# = (sqrt2-sqrtx)/(sqrt(2x)(x-2))#

# = (-(sqrtx-sqrt2))/(sqrt(2x)(sqrtx-sqrt2)(sqrtx+sqrt2))#

# = (-1)/(sqrt(2x)(sqrtx+sqrt2))#

So,

#lim_(xrarr2) (1/sqrtx-1/sqrt2)/(x-2) =lim_(xrarr2) (-1)/(sqrt(2x)(sqrtx+sqrt2))#

# = (-1)/(sqrt4(sqrt2+sqrt2)) = (-1)/(4sqrt2) = - sqrt2/8#

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