Question

I need clarification on how to find a limit. My problem is:

"lim 4th root of (9 - x^2), as x approaches 5"

*Sorry- my iPad's being gay, and I can't type it any nicer than that... :/

So... When I plug in 5 for x, I get the 4th root of -16, which, since 4 is even, would still be a negative... Which is irrational/ not possible right? So... What's the limit for this one?
Any and all help is greatly appreciated! :)

Answer 1

@wolfe8 Help? :)

Answer 2

Any ideas?

Answer 3

To be honest I forgot how to do this.

Answer 4

And it's too late for me to figure it out. Sorry. Good luck.

Answer 5

*Rage!* haha! okay... Thanks for taking a look :)

Answer 6

I'll keep it open, ya know, in case you feel inspired later ;)

Answer 7

And for other OS helpers.

Answer 8

\[
\lim_{x\to 5} \, \sqrt[4]{9-x^2}=(1+i) \sqrt{2}
\]
The answer is a complex number
http://www.wolframalpha.com/input/?i=limit+%28+9-x^2%29^%281%2F4%29+when+x+goes+to+5

Answer 9

So (1 + i)sqrt2 is the limit? That's the value?

Answer 10

What's what I looked for too, but I just don't get the x=5 part

Answer 11

Yes

Answer 12

Ahhh ok I see what I did wrong now. Yea that should be right.

Answer 13

I still don't understand what it all means, like how the graph is still existing & imaginary... IDK... I'm not too hot on limits in general (We all have that one chapter, aye?) but I can usually at least find them... *sigh* but I'm going to study the graph more, and at least I have something for my paper :) Thank you guys!

Answer 14

Any insight @linh412986 ?

Answer 15

Nothing, in your case, just plug in x = 5, then you have \[\sqrt[4]{-16} = 2 \sqrt[4]{-1} = 2 \sqrt[4]{i^2}\]
\[= 2i^{\frac{1}{2}}\]
with i^2 = -1
The answer should be an imaginary number.