Solve: Lim $x \to 1$ $x->1$ $\frac{\sqrt{x} - \sqrt{2 - x^{2}}}{2 x - \sqrt{2 + 2 x^{2}}}$ $(sqrt(x)-sqrt(2-x^2))/(2x-sqrt(2+2x^2)$ ?

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Lim $x \to 1$ $x->1$ $\frac{\sqrt{x} - \sqrt{2 - x^{2}}}{2 x - \sqrt{2 + 2 x^{2}}}$ $(sqrt(x)-sqrt(2-x^2))/(2x-sqrt(2+2x^2)$

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Answer 1 · 2018-03-20T19:45:33

$\frac{3}{2}$ $3/2$

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

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

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

= $1/2((x+2)/(x+1))$ 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

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

= $1/2*3/2*4/2=3/2$