Question

How do you evaluate `lim_(xtooo) (5 - x^(1/2))/(5 + x^(1/2))` ?

Answer 1

Use L'Hôpital's rule

Explanation:

Given: `lim_(xtooo) (5 - x^(1/2))/(5 + x^(1/2))`

Because the expression evaluated at the limit yields the indeterminate form `oo/oo`, the use of L'Hôpital's rule is warranted.

You apply the rule by differentiating the number and the denominator:

`lim_(xtooo) ((d((5 - x^(1/2))))/dx)/((d((5 + x^(1/2))))/dx) =`

`lim_(xtooo) (-1/2x^(-1/2))/(1/2x^(-1/2)) =`

`lim_(xtooo) -1 = -1`

According to the rule, so goes the original limit:

`lim_(xtooo) (5 - x^(1/2))/(5 + x^(1/2)) = -1`

Answer 2

I got: `-1`

Explanation:

If you try directly you get:

`lim_(x->oo)(5-x^(1/2))/(5+x^(1/2))=-oo/oo`

We can use de L'Hospital Rule (deriving top and bottom to get):

`lim_(x->oo)(5-x^(1/2))/(5+x^(1/2))=lim_(x->oo)(-1/2cancel(x^(-1/2)))/(1/2cancel(x^(-1/2)))=-(1/2)/(1/2)=-1`