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

Expert Answers

An illustration of the letter 'A' in a speech bubbles

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.

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial