What is the limit as Δχ approaches 0 for the equation f(x) = 1/√x with the difference quotient applied?,

Question

  lim       1

             —

Δχ→0    √χ

Answer 1

Using h instead of Δx.

As a difference quotient:

lim_h->0[(1/√(x+h)) - 1/√x]/h (multiply every term, top and bottom, by √x √(x+h)); leaving off the lim for now.

[√x - √(x+h)] / [ h √x √(x+h)]

(multiply numerator and denominator by the conjugate of the numerator)

[( √x - √(x+h) ) ( √x + √(x+h)] / [ h √x √(x+h) (√x + √(x+h)]

Difference of squares in the numerator:

[ x - (x + h)] / [ h √x √(x + h) (√x + √(x+h))]

-h / [h √x √(x+h) (√x + √(x+h))]

The "h" cancels as h approaches zero, but it not equal to zero:

-1 / [ √x √(x+h) (√x + √(x+h))]

Take the limit as h->0:

-1 / [√x √x (√x + √x)]

-1 / [x (2√x) = -1 /(2x√x) or -1 /(2x^3/2)

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