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gillian88 · Jan 28, 2006, 02:07 PM

For what value(s) of k does the limit lim x ->3 (√x) - k / x - 3 exist?

thanks =]

reinsuranc · Jan 28, 2006, 02:53 PM

Hi,

You need to do a little better job in this forum with making the problem clear regarding where the square root sign ends and where the denominator ends.

I think you are asking:

for what values of k does the limit
as x->3 of [(x-k)^.5]/(x-3) exist.

Assuming I have correctly restated the problem, the only time the limit of this expression will not exist is if the denominator is zero, which occurs when x=3.

If x=3, the limit does not exist for all values of k. If k also equals 3, the expression reduces to 1/[(x-3)^.5], which is fine everywhere except when x=3.

If x does not equal 3, the limit exists for all values of k.

gillian88 · Jan 28, 2006, 05:11 PM

Sorry, only the 3 is square-rooted

CroCivic91 · Jan 28, 2006, 05:44 PM

Now restate the problem using parentheses.

Is it:
(lim x -> sqrt(3)) (x - k) / (x - 3)
or
(lim x -> sqrt(3)) x - (k / (x - 3))
or
(lim x -> sqrt(3)) ((x - k) / x) - 3
or
(lim x -> sqrt(3)) x - (k / x) - 3
or
(lim x -> 3) (x - k) / (x - sqrt(3))
or
...

You get the point..?

gillian88 · Jan 29, 2006, 01:17 PM

here.. itz like this..

if u can't see the image then go to http://img.photobucket.com/albums/v3...mitproblem.jpg

limit goes to 3 ((sqrt x) - k) / (x - 3)

(sqrt x only!)