# Determine the limit and where possible compute.

• Nov 2nd 2009, 03:22 PM
cpj
Determine the limit and where possible compute.
hi, the problem i have is i'm having trouble with how to determine the limit using first principles, any help would be appreciated. the problem is..

lim (5x-2)^1/2 - 2(2^1/2)
x->2 (x^2)-(3x)+2

thanks for any help, i've never had much luck with limits
• Nov 2nd 2009, 03:55 PM
mr fantastic
Quote:

Originally Posted by cpj
hi, the problem i have is i'm having trouble with how to determine the limit using first principles, any help would be appreciated. the problem is..

lim (5x-2)^1/2 - 2(2^1/2)
x->2 (x^2)-(3x)+2

thanks for any help, i've never had much luck with limits

Multiply numerator and denominator by the conjugate surd $\displaystyle \sqrt{5x - 2} + \sqrt{8}$, simplify the numerator, factorise the denominator, cancel the common factor and then take the limit. Please show your work if you need more help.
• Nov 3rd 2009, 09:38 AM
cpj
(5x-2)^1/2 - (8)^1/2
x^2 - 3x + 2

[(5x-2)^1/2 - (8)^1/2] * [(5x-2)^1/2 + (8)^1/2]
[(5x-2)^1/2 + (8)^1/2]*[(x^2)-3x+2]

(5x-2)-8
[(5x-2)^1/2 + (8)^1/2]*(x-1)(x-2)

5
-2(2^1/2)

this is my final limit, could you tell me if its correct please. i'm not sure..
• Nov 3rd 2009, 09:45 PM
mr fantastic
Quote:

Originally Posted by cpj
(5x-2)^1/2 - (8)^1/2
x^2 - 3x + 2

[(5x-2)^1/2 - (8)^1/2] * [(5x-2)^1/2 + (8)^1/2]
[(5x-2)^1/2 + (8)^1/2]*[(x^2)-3x+2]

(5x-2)-8
[(5x-2)^1/2 + (8)^1/2]*(x-1)(x-2)

5
-2(2^1/2)

this is my final limit, could you tell me if its correct please. i'm not sure..

I get $\displaystyle \frac{5}{2 \sqrt{8}}$.