Limit x approches 1(sqrt(1+2x)-(sqrt3x))/(sqrt(3+x)-2sqrt(x))

Answer 1

lim (x->1) (√(1+2x)-√(3x))/(√(3+x)-2√x)

The fraction at x=1 is 0/0, so we have to try to resolve that. Multiplying by the conjugate of the denominator we have

(√(1+2x)-√(3x))/(√(3+x)-2√x)) * (√(3+x)+2√x)/(√(3+x)+2√x)

(√(1+2x)-√(3x))*(√(3+x)+2√x) / (3+x - 4x)

as x->1 that becomes

((√3-√3)(√3+2))/(3+1-4)

Hmmm. STill 0/0. I guess we have to try l'Hospital's Rule. Taking derivatives top and bottom we get

1/√(1+2x) - √3/(2√x)
-------------------------
1/(2√(3+x) - 1/√x

-> (1/√3 - √3/2)/(1/2√4 - 1/√1)
= 2/(3√3)