How would you expand the brackets of (2x+2Δx +1)^1/2?
Update:
Thanks for your reply. The whole question is:
How do you calculate the limit of 2/(√(2*x+2Δx +1)+√(2*x+1)) as Δx->0?
Update 3:EDIT. SEE ABOVE
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i) Generally in maths, Δx is considered to be tending to zero.
ii) So, grouping this {(2x + 1) + 2Δx}^(1/2) = {√(2x + 1)}*[1 + 2h/(2x + 1)]^(1/2),
where h = 2Δx
iii) The expansion of this =
= {√(2x + 1)}*[1 + (1/2)*{2h/(2x + 1)} + {(1/2)(1/2 - 1)/2!}*{2h/(2x + 1)}² + --------]
= {√(2x + 1)}*[1 + {h/(2x + 1)} - {(1/2!}*{h/(2x + 1)}² + --------]
EDIT:
I just after about 21 hours, has noticed your additional details.
As you just needed expansion the same is presented. If you need the final answer as 2, then there must be some conditions given. Hence kindly place your question in full, then we can provide the solution accordingly.
expand as : (a+b+c)2 = a2 +b2 + c2+ 2(ab +ac+ bc); put a=2x, b= 2deltax, c= 1, & simplify !
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