Question

Find limit?

Find (in terms of the constant aa)
`lim_(h→0)(√4(a+h) - √4a )/ h`

Limit =
equation editor

Answer 1

`"The Reqd. Lim.="1/sqrta`.

Explanation:

`"The Reqd. Lim.="lim_(h to 0){sqrt(4(a+h))-sqrt(4a)}/h`,

`=lim{2sqrt(a+h)-2sqrta}/h`,

`=2lim[{sqrt(a+h)-sqrta}/hxx1]`,

`=2lim[{sqrt(a+h)-sqrta}/hxx{sqrt(a+h)+sqrta}/{sqrt(a+h)+sqrta}]`,

`=2lim{(sqrt(a+h))^2-(sqrta)^2}/{h(sqrt(a+h)+sqrta)}`,

`=2lim{a+h-a}/{h(sqrt(a+h)+sqrta)}`,

`=2lim_(h to 0)1/{(sqrt(a+h)+sqrta)}`,

`=2*1/(sqrta+sqrta)`.

`rArr "The Reqd. Lim.="1/sqrta`.

Answer 2

`Lim/(h->0) (sqrt(4a)) = 1/(2sqrt(4a))`

Explanation:

Corrected from previous edition.

You already have half of the work completed by plugging f(x) into the difference quotient.

We have a domain restriction that we need to find a way to algebraically manipulate out of our function. We can't plug 0 into the denominator, so let's find a way to get rid of it.

`Lim/(h->0) (sqrt(4a)) = Lim/(h->0) ((sqrt(4(a+h))-(sqrt(4a)))/h)`

The next step would be to multiply the top and bottom by the conjugate to clear the square root functions in the numerator.

`=Lim/(h->0) ((sqrt(4a+4h))-(sqrt(4a)))/h * ((sqrt(4a+4h)+(sqrt(4a)))/(sqrt(4a+4h)+(sqrt(4a))))`

Next, we simplify the functions by combining like terms in the numerator. Don't worry about the denominator, in most cases you will leave it in factored form.

`=Lim/(h->0) (4a+4h-4a)/(h*(sqrt(4a+4h)+(sqrt(4a))`

In this case, our 4a and -4a will sum to zero and we will be left with h in the numerator

`=Lim/(h->0) (4h)/(h*(sqrt(4a+4h)+(sqrt(4a))`

Our h factor will divide out of the numerator and denominator and we will be left with the following.

`=Lim/(h->0) (4)/((sqrt(4a+4h)+(sqrt(4a))`

Now that our domain restriction has been removed, we can proceed with the direct substitution portion of the limit. Sub-in 0 for h and evaluate.

`f`(x) = (4)/((2*2)sqrt(a)`

What we're left with is:

`f`(x) = (1)/(sqrt(a)`