## How does one solve this?

For the discussion of math. Duh.

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pietertje
Posts: 97
Joined: Sat Nov 08, 2008 12:43 pm UTC

### How does one solve this?

I'm having a bit of trouble with this one, how do you solve this?

EDIT: The "= 0" isn't supposed to be there.

Mindworm
Posts: 88
Joined: Wed Sep 22, 2010 4:06 pm UTC
Location: The dark place where the counterexamples live.

### Re: How does one solve this?

I don't see the problem. It's defined at 2 and in the neighbourhood (between 0 and 2, which is where your limit comes from), where it behaves nicely.

Note that the limit is also 0 from the other side (in complex numbers, that is), no matter what branch of the square root you look at.
Any reason why this should not be 0?
The cake is a pie.

Hix
Posts: 364
Joined: Sun Oct 15, 2006 5:46 pm UTC

### Re: How does one solve this?

Theorem: The limit of a product of two terms is the product of the limits of those two terms (assuming that those two limits exist)

voidPtr
Posts: 140
Joined: Sun Apr 26, 2009 6:53 pm UTC

### Re: How does one solve this?

Hi,

Are you having difficulty with limits in general or just this question specifically? The reason I ask is because as far as these kinds of questions go, this one is as straightforward as it gets. Given *any* limit question it's helpful to first try and plug in the value and see what you get. Is the function defined at that value? (Yes.) Is the function continuous on the interval containing that value? (Yes.) It's as easy as middle-school algebra if the answer to both of those questions is yes.

crazydave
Posts: 6
Joined: Mon Nov 02, 2009 9:30 pm UTC

### Re: How does one solve this?

voidPtr wrote:Hi,

Are you having difficulty with limits in general or just this question specifically? The reason I ask is because as far as these kinds of questions go, this one is as straightforward as it gets. Given *any* limit question it's helpful to first try and plug in the value and see what you get. Is the function defined at that value? (Yes.) Is the function continuous on the interval containing that value? (Yes.) It's as easy as middle-school algebra if the answer to both of those questions is yes.

I could be mistaken, but I do not think that the function is continuous on the interval containing that value. Once when x is bigger than 2, the equation under the radical is always negative, thus yielding a complex number. This problem sounds like a problem from the limit section of a calculus 1 course, so I would imagine that we are talking about the reals not the complex numbers.

What might be causing you confusion on this problem is the x->2-. This is just directing you to which way to take the limit from. The negative sign means that you need to be approaching 2 from the left, or in other words from values less than 2. A lot of the time it doesn't matter which way we approach a value from, but in this problem we need to tell which way to take the limit because the function is not defined for numbers greater than 2. Note the domain is [-2,2]

Similarly, x->2+ would mean approaching 2 from right.

D.B.
Posts: 197
Joined: Wed Sep 09, 2009 3:08 pm UTC

### Re: How does one solve this?

I'm sure someone can correct me here, but the way I'd go about doing it would be to define
$x = 2 - \delta$
and take the limit as [imath]\delta \rightarrow 0^+[/imath] so we get
$\lim_{\delta \rightarrow 0^+} (2 - \delta)\sqrt{4 - (2 - \delta)^2}$
We expand out the squared term
$\lim_{\delta \rightarrow 0^+} (2 - \delta)\sqrt{\delta(4 - \delta)}$
And then it's easy. [imath](2 - \delta)[/imath] does nothing special, [imath]\sqrt{\delta(4 - \delta)}[/imath] behaves similarly, and the extra factor of [imath]\sqrt{\delta}[/imath] tends towards zero.

If we let [imath]x \rightarrow 2^+[/imath] then it's slightly different as we now get

$x = 2 + \delta$
$\lim_{\delta \rightarrow 0^+} (2 - \delta)\sqrt{4 - (2 + \delta)^2}$
$\lim_{\delta \rightarrow 0^+} (2 - \delta)\sqrt{-\delta(4 + \delta)}$
which has an extra factor of [imath]i[/imath] in it but will still tend to zero.