Question

the limit of [ sqrt(3+x) - sqrt3 ] / x as x approaches 0?

Answer 1

why is the answer not zero?

Answer 2

because although you get a zero in the numerator, you also get a zero in the denominator

Answer 3

the trick to doing this is to rationalize the numerator, so you can cancel the \(x\) top and bottom
multiply by
\[\frac{\sqrt{x+3}+\sqrt{x}}{\sqrt{x+3}+\sqrt{x}}\] leave the denominator in factored form

Answer 4

how did you get there from the original equation?

Answer 5

multiply by the conjugate

Answer 6

conjugate of \(\sqrt{a}+\sqrt{b}\) is \(\sqrt{a}-\sqrt{b}\) and this gimmick works because \((\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})=a-b\)

Answer 7

oh maybe i confused you
i did not say that was the original equation, i said that is what you need to multiply the original equation by
you should write
\[\frac{\sqrt{x+3}-\sqrt{x}}{x}\times \frac{\sqrt{x+3}+\sqrt{x}}{\sqrt{x+3}+\sqrt{x}}\]

Answer 8

ok i see i have made a mistake it should be this
\[\frac{\sqrt{x+3}-\sqrt{3}}{x}\times \frac{\sqrt{x+3}+\sqrt{3}}{\sqrt{x+3}+\sqrt{3}}\]

Answer 9

the second term is \(\sqrt{3}\) not \(\sqrt{x}\)

Answer 10

then the numerator will be \(x+3-3=x\) which you then cancel with the \(x\) in the denominator to leave
\[\frac{1}{\sqrt{x+3}+\sqrt{x}}\] and then you replace \(x\) by 0 to get your result

Answer 11

ohhh okay
thank youuu

Answer 12

yw