# Evaluate lim x---2 (Square root 6-x) -2 over (square root 3-x)-1 (it's a rational function).

The variable x tends to 2 (or to -2). The function is `f ( x ) = ( sqrt ( 6 - x ) - 2 ) / ( sqrt ( 3 - x ) - 1 ),` and it is not rational because it contains square roots. But it is not an obstacle to find the limit.

Both numerator and denominator are continuous functions near `x = 2` and near `x = -2 .`

If `x` tends to `-2 ,` then the numerator tends to `sqrt ( 6 + 2 ) - 2 = 2 sqrt ( 2 ) - 2` and the denominator tends to `sqrt ( 3 + 2 ) - 1 = sqrt ( 5 ) - 1 ,` which is nonzero. Therefore, by the theorem of a quotient limit the limit is, well,

`( 2 sqrt ( 2 ) - 2 ) / ( sqrt ( 5 ) - 1 ) = 2 (sqrt(2)-1)(sqrt(5)+1)/ (5-1) = ((sqrt(2)-1)(sqrt(5)+1))/2 approx 0.064 .`

If `x` tends to `+2 ,` then the numerator tends to `sqrt ( 6 - 2 ) - 2 = sqrt( 4 ) - 2 =0,` and the denominator tends to `sqrt ( 3 - 2 ) - 1 = sqrt ( 1 ) - 1 =0,` so we have an indeterminacy `0/0.` To resolve it, multiply numerator and denominator by both conjugate root expressions:

`f ( x ) = (( sqrt ( 6 - x ) - 2 )( sqrt ( 6 - x ) +2 )( sqrt ( 3 - x ) + 1 )) / (( sqrt ( 3 - x ) - 1 )( sqrt ( 3 - x ) + 1 )( sqrt ( 6 - x ) + 2 )) =` `=(( sqrt ( 6 - x ) - 2 )( sqrt ( 6 - x ) +2 )) / (( sqrt ( 3 - x ) - 1 )( sqrt ( 3 - x ) + 1 )) *( sqrt ( 3 - x ) + 1 )/(sqrt ( 6 - x ) + 2 ) =`
`=(6-x-4)/(3-x-1) * ( sqrt ( 3 - x ) + 1 )/(sqrt ( 6 - x ) + 2 ) =`
`=(2-x)/(2-x) * ( sqrt ( 3 - x ) + 1)/(sqrt ( 6 - x ) + 2 ) =( sqrt ( 3 - x ) + 1 )/(sqrt ( 6 - x ) + 2 ).`

Now both parts have nonzero limits, and we may apply the theorem of a quotient limit: `f(x)->(1+1)/(2+2)` = 1/2.