Can you help me solve this limit (calculus) please?

Question

I need to evaluate (aka solve) the limit --

[ (square root of 6-x) - 2 ] / [ (square root of 3-x) - 1 ]

as x approaches 2. I've been trying to rationalize the square roots and/or factoring, but I keep getting stuck. I'm pretty sure the limit exists because when I graphed it (note: on my calculator), the (y)value of x approaching 2 came out to around 0.48 or a value less than 1.

Answer 1

[x→2]lim (√(6-x)-2)/(√(3-x)-1)

First, rationalize both numerator and denominator by multiplying both numerator and denominator by (√(6-x)+2)(√(3-x)+1), then simplify and evaluate:

[x→2]lim (√(6-x)-2)(√(6-x)+2)(√(3-x)+1) / ((√(3-x)-1)(√(6-x)+2)(√(3-x)+1))

[x→2]lim (√(3-x)+1)/(√(6-x)+2) * (6-x-4)/(3-x-1)

[x→2]lim (√(3-x)+1)/(√(6-x)+2) * (2-x)/(2-x)

[x→2]lim (√(3-x)+1)/(√(6-x)+2)

(√(3-2)+1)/(√(6-2)+2)

(√1 + 1)/(√4 + 2)

1/2

Answer 2

use L-Hospitals rule.
In case u don't know what it is :
whenever there is 0/0 or infinity/infinity form in a limit, than u can rationalise the denominator & numerator seperately & then put the limit.
ur ans will be 1/2.