Calculus Question (regarding limits)

[Chris418]

Hey,

I dont think im supposed to put this here, but im in dire need of help (i took calc 5 years ago).

lim (x->9) f(x)=(9-x)/(3-(x)^.5)

Any ideas... i know the answer is 6, just not sure how to derive it.

Thanks in advance.

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[BellyDancingDoc]

Ummm, no homework quesitons allowed in the Pre-Allo forum.

But for whatever it's worth, some folks are petitioning to have a Pre-med Homework forum opened. Try doing a search for the discussion and then (nicely) pestering a mod...

 

[Chris418]

eh, its not really a homework question; i just can't figure it out and im getting irritated... my bad though.

 

[Stroganoff]

Plug 9 in and you get the indeterminate 0/0, so then you use L'Hopital's rule!

(Are you seeing that it's 9-9=0 in the numerator and 3-√9=0 in the denominator?)

 

[Chris418]

Alright, awesome... thanks, i appreciate it.

 

[pyrois]

Hey,

I dont think im supposed to put this here, but im in dire need of help (i took calc 5 years ago).

lim (x->9) f(x)=(9-x)/(3-(x)^.5)

Any ideas... i know the answer is 6, just not sure how to derive it.

Thanks in advance.

Sigh, I can't leave a fellow pre-medder in such peril

There's a whole bunch of ways to do the problem. The most direct approach is simply multiplying top and bottom by 3 + (x)^0.5

The denominator will become (9-x) which will cancel with the top and leave 3 + (x)^0.5 which is 6 when you plug in 9.

Alternatively you can do The French Hospital's rule (L'Hospital) and differentiate top/bottom.

*edit* somebody ninja'd my response oh well.

 

[braluk]

No homework questions here.