Solve: Lim #x->1# #(sqrt(x)-sqrt(2-x^2))/(2x-sqrt(2+2x^2)# ?

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Solve:

Lim #x->1# #(sqrt(x)-sqrt(2-x^2))/(2x-sqrt(2+2x^2)#

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Answer 1 · 2018-03-20T19:45:33

#3/2#

#lim_(x->1)(sqrt(x)-sqrt(2-x^2))/(2x-sqrt(2+2x^2)#

#(sqrt[x] - sqrt[2 - x^2])/(2 x - sqrt[2 + 2 x^2]) *(sqrt[x] + sqrt[ 2 - x^2])/(2 x + sqrt[2 + 2 x^2])#

= #(x-2+x^2)/(4x^2-2-2x^2)=((x+2)(x-1))/(2(x+1)(x-1))#

= #1/2((x+2)/(x+1))# and then

#(sqrt(x)-sqrt(2-x^2))/(2x-sqrt(2+2x^2)) = 1/2((x+2)/(x+1))(2 x + sqrt[2 + 2 x^2])/(sqrt[x] + sqrt[ 2 - x^2])#

and then

#lim_(x->1)(sqrt(x)-sqrt(2-x^2))/(2x-sqrt(2+2x^2)) = lim_(x->1) 1/2((x+2)/(x+1))(2 x + sqrt[2 + 2 x^2])/(sqrt[x] + sqrt[ 2 - x^2])#

= #1/2*3/2*4/2=3/2#