Question

`limxrarra (x^2-a^2)/(sqrtx-sqrta)=32` What value should "a" have?

Answer 1

The value of `a=4`

Explanation:

To calculate the limit

`lim_(x->a)(x^2-a^2)/(sqrtx-sqrta)=(a^2-a^2)/(sqrta-sqrts)=0/0`

Which is an an indeterminate form

We apply l'Hôpital's rule

`lim_(x->a)(x^2-a^2)/(sqrtx-sqrta)=lim_(x->a)((x^2-a^2)')/((sqrtx-sqrta)')`

`=lim_(x->a)(2x)/(1/(2sqrtx))`

`=4asqrta`

Therefore,

`4asqrta=32`

`a^(3/2)=32/4=8`

`a=8^(2/3)`

`a=2^2=4`

`a=4`

Answer 2

`a=4`

Explanation:

We have:

`L = lim_(x rarr a) (x^2-a^2)/(sqrtx-sqrta) = 32`

Which we can write as:

`L = lim_(x rarr a) (x^2-a^2)/(sqrtx-sqrta) * (sqrtx+sqrta)/(sqrtx+sqrta)`
`\ \ \ = lim_(x rarr a) ((x+a)(x-a))/(x-a) * (sqrtx+sqrta)`
`\ \ \ = lim_(x rarr a) (x+a) (sqrtx+sqrta)`
`\ \ \ = (a+a) (sqrta+sqrta)`
`\ \ \ = (2a) (2sqrta)`
`\ \ \ = 4a^(3/2)`

So if:

`L = 32 => 4a^(3/2) =32`
`:. a^(3/2) =8`
`:. a \ \ = 4`

Answer 3

`a=4`

Explanation:

Factor as a difference of squares multiple times. You may want to recall that `u^2-v^2=(u+v)(u-v)`. Expand the right hand side to confirm this equality. Using this identity:

`lim_(xrarra)(x^2-a^2)/(sqrtx-sqrta)=lim_(xrarra)((x+a)(x-a))/(sqrtx-sqrta)`

We'll again use `u^2-v^2=(u+v)(u-v)`, but here `u=sqrtx` and `v=sqrta`.

`=lim_(xrarra)((x+a)((sqrtx)^2-(sqrta)^2))/(sqrtx-sqrta)`

`=lim_(xrarra)((x+a)(sqrtx+sqrta)(sqrtx-sqrta))/(sqrtx-sqrta)`

Now the `sqrtx-sqrta` terms cancel one another out:

`=lim_(xrarra)(x+a)(sqrtx+sqrta)`

`=lim_(xrarra)(a+a)(sqrta+sqrta)`

`=(2a)(2sqrta)`

`=4a^(3/2)=32`

Thus:

`a^(3/2)=8`

`a=(2^3)^(2/3)=2^2=4`