find lim ∆x--> 0 ... [ ( √ x + ∆x ) - ( √ x ) ] / ∆x?

Question

in the question, the x and the delta x are all under the root

Answer 1

Multiply the numerator and denominator by (√ x + ∆x ) + ( √ x )
[(√ x + ∆x ) - ( √ x )][( √ x + ∆x ) + ( √ x )]
= x + ∆x - x
= ∆x
∆x/[∆x[(√ x + ∆x ) + ( √ x )]]
= 1/[(√ x + ∆x ) + ( √ x )]
Replace ∆x with 0.
= 1/[(√ x + 0 ) + ( √ x )]
= 1/[(√ x ) + ( √ x )]
= 1/(2√ x )

Answer 2

The answer is the first derivative of √ x, which is equal to 1/(2*√ x).

To solve the problem, first we try by directly applying the limits ∆x -> 0, but this gives us a 0/0 answer, which means we have to do some further simplification.

We notice that the numerator term is causing the problem, so we simplify it by first eliminating the square root & then eliminating the ∆x term. To do that, we multiply numerator & denominator by the conjugate term of the numerator, which will be ( √ x + ∆x ) + ( √ x ), therefore our problem works out to be:

lim ∆x->0 { (√ x + ∆x ) - ( √ x ) } * { (√ x + ∆x ) + ( √ x ) } / [ ∆x * { (√ x + ∆x ) + ( √ x ) } ] =

lim ∆x->0 { (√ x + ∆x )^2 - ( √ x )^2 } / [ ∆x * { (√ x + ∆x ) + ( √ x ) } ] =

lim ∆x->0 { x + ∆x - x } / [ ∆x * { (√ x + ∆x ) + ( √ x ) } ] =

lim ∆x->0 [ ∆x / [ ∆x * { (√ x + ∆x ) + ( √ x ) } ] ] =

lim ∆x->0 [ 1 / { (√ x + ∆x ) + ( √ x ) } =

Now we have a form where we can conveniently apply the limits as ∆x->0, giving us:

= [ 1 / ( √ x + √ x ) ]

= 1 / (2 * √ x )

This is the answer

Answer 3

1+1=two points

Answer 4

Do a web search for L'Hopitals rule

Answer 5

hint: multiply by conjugate of numerator...

Answer 6

See this question and answers:

http://answers.yahoo.com/question/index?qid=20060606164852AARgVX4

Answer 7

dont know how