Question

Question #d2448

Answer 1

`=- 1/2 1/(x^(3/2))`

Explanation:

So you are here?

`lim_(h to 0) 1/h (1/sqrt(x+h)-1/sqrt(x))`

If so, you can now add the terms:

`=lim_(h to 0) 1/h ((sqrt x - sqrt (x+h)) /(sqrt x sqrt(x+h)))`

And here is the trick: multiply top and bottom of this resulting fraction by the conjugate of the numerator:

`=lim_(h to 0) 1/h (((sqrt x - sqrt (x+h)) * (sqrt x + sqrt (x+h))) /(sqrt x sqrt(x+h)(sqrt x + sqrt (x+h))))`

`=lim_(h to 0) 1/h ( x - (x+h) ) /(sqrt x sqrt(x+h)(sqrt x + sqrt (x+h))))`

`=lim_(h to 0) 1/h (( -h) /(sqrt x sqrt(x+h)(sqrt x + sqrt (x+h))))`

`=- lim_(h to 0) ((1) /(sqrt x sqrt(x+h)(sqrt x + sqrt (x+h))))`

`=- ((1) /(sqrt x sqrt(x)(sqrt x + sqrt (x))))`

`=- ((1) /( sqrt x * sqrt(x) * 2 sqrt x ))`

`=- 1/2 1/(x^(3/2))`

Answer 2

Combine the quotients into a single quotient.

Explanation:

`1/sqrt(x+h) - 1/sqrtx = (sqrtx-sqrt(x+h))/(sqrtxsqrt(x+h))`

My guess (educated guess) is that your textbook/instructor lists the 3rd step as

Find `(f(x+h)-f(x))/h`.

If so, then our step 3 for this function is

`(f(x+h)-f(x))/h = ((sqrtx-sqrt(x+h))/(sqrtxsqrt(x+h)))/h`

`= ((sqrtx-sqrt(x+h))/(sqrtxsqrt(x+h)))/(h/1)`

`= ((sqrtx-sqrt(x+h))/(sqrtxsqrt(x+h))) * (1/h)`

`= (sqrtx-sqrt(x+h))/(hsqrtxsqrt(x+h))`

Now remove the radicals in the numerator ("rationalize the numerator")

`= ((sqrtx-sqrt(x+h)))/(hsqrtxsqrt(x+h)) * ((sqrtx+sqrt(x+h)))/((sqrtx+sqrt(x+h)))`

`= (x-(x+h))/(hsqrtxsqrt(x+h)(sqrtx+sqrt(x+h))`

`= (-h)/(hsqrtxsqrt(x+h)(sqrtx+sqrt(x+h))`

`= (-1)/(sqrtxsqrt(x+h)(sqrtx+sqrt(x+h))`

And Step 4 evaluate the limit as `hrarr0` -- sometimes described as "plug in `0` for `h`"

`lim_(hrarr0)(-1)/(sqrtxsqrt(x+h)(sqrtx+sqrt(x+h))) = (-1)/(sqrtxsqrt(x+0)(sqrtx+sqrt(x+0)))`

`= (-1)/(sqrtx sqrtx(2sqrtx))`

`= (-1)/(2sqrtx^3)`

Note

As you can see from the other answers, many treatments of calculus do not use a multi-step description of finding `lim_(hrarr0)(f(x+h)-f(x))/h`.

The other presentations all do the same steps, they just don't bother with separating the steps.

Answer 3

Sorry, I misunderstood your question.

Explanation:

Here is how you solve it using the limit definition:

`lim_(h->0) (f(x+h) - f(x))/h`

`lim_(h->0) (1/sqrt(x+h) - 1/sqrt(x))/h`

`lim_(h->0) 1/(sqrt(x+h)*h) - 1/(sqrt(x)*h)`

`lim_(h->0) sqrt(x)/(sqrt(x+h)*h*sqrt(x)) - sqrt(x+h)/(sqrt(x)*h*sqrt(x+h))`

`lim_(h->0) (sqrt(x)-sqrt(x+h))/(sqrt(x)*h*sqrt(x+h))`

`lim_(h->0) (x-(x+h))/ (sqrt(x)*h*sqrt(x+h)*(sqrt(x)+sqrt(x+h))`

`lim_(h->0) (-h)/ (sqrt(x)*h*sqrt(x+h)*(sqrt(x)+sqrt(x+h))`

`lim_(h->0) (-1)/ (sqrt(x)*sqrt(x+h)*(sqrt(x)+sqrt(x+h))`

plug in h = 0 because it can now be evaluated

=`-1/2x^(-3/2)`

Answer 4

`g'(x)=-1/2x^(-3/2)`

Explanation:

It appears from what is written that you want to obtain the derivative from`color(blue)" limit definition"`

`color(red)(bar(ul(|color(white)(2/2)color(black)(f'(x)=lim_(hto0)(f(x+h)-f(x))/h)color(white)(2/2)|)))`
the aim of the exercise to eliminate h from the denominator.

`f(x)=1/sqrtx`

`rArrf'(x)=lim_(hto0)(1/sqrt(x+h)-1/sqrtx)/h`

At this stage, rationalise the numerator by multiplying numerator/denominator by the `color(blue)"conjugate"` of the numerator.

`rArrlim_(hto0)((1/sqrt(x+h)-1/sqrtx)(1/sqrt(x+h)color(red)(+)1/sqrtx))/(h(1/sqrt(x+h)color(red)(+)1/sqrtx)`

`=lim_(hto0)((1/(x+h)-1/x))/(h(1/sqrt(x+h)+1/sqrtx)`

`=lim_(hto0)(((x-x-h))/(x(x+h)))/(h(1/sqrt(x+h)+1/sqrtx)`

`=lim_(hto0)-cancel(h)^1/(x(x+h))xx1/(cancel(h)^1(1/sqrt(x+h)+1/sqrtx)`

`=(-1)/(x^2(1/sqrtx+1/sqrtx))=(-1)/(x^2(2/sqrtx)`

`=(-1)/((2x^2)/x^(1/2))=(-1)/(2x^(3/2))=-1/2x^(-3/2)`

`rArrg'(x)=-1/2x^(-3/2)larr" from first principles"`