Limit = 0, or does it exist?

May 2010
22
0
I have a problem: \(\displaystyle \lim_{x \to -1} f(x) =\sqrt{x} * (x+1)\)

I enter this into my TI-89 and get 0. The book says the limit does not exist. How do I figure out why? Is it because I cannot put a -1 in the radical?

Thanks.
 
Last edited:
May 2010
22
10
sqrt(x) is defined for x positive. so you cannot search for its limit at x=-1
 
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Jan 2010
354
173
The TI-89 is actually correct, but some clarification is needed.

In the domain of the real numbers, the function is undefined around \(\displaystyle x=-1\) and therefore the limit is undefined.

However, in the complex plane, the function is defined around \(\displaystyle x=-1\) and in fact the limit at \(\displaystyle x=-1\) is equal to 0. This is why it gives that result. However, you are not expected to treat the function this way so you should disregard the result.
 
May 2010
22
0
Getting there....

That clarifies it for the most part, but since I have the TI-89 in Real mode, it seems like it should un-define all imaginary numbers.
 
Jan 2010
354
173
Yes, it would be nice if it worked that way, but what that mode really does is just return an error any time a non-real answer is obtained. Since 0 is a real number, it doesn't give an error.
 
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May 2010
22
0
Great

Now it makes sense. Awesome! So from now on, I should know that the limit is undefined as x approaches a negative number, with x in a radical in the function. Is this correct?
 
Jan 2010
354
173
More or less. The proper question you should ask is this: "Is the function defined immediately to the left and right of the point where the limit is taken?"

So for example, with your original problem, you might ask yourself: "Is the function defined at \(\displaystyle x=-1.000001\)? Is the function defined at \(\displaystyle x=-0.999999\)?"

If either answer is no, then the limit doesn't exist at that point.

If both answers are yes, it's possible that the limit exists, but you need to check further to see if it actually does.

In general, like you concluded, it will work out that any function that contains \(\displaystyle \sqrt{x}\) will not have a limit at any negative number. However consider another example where a function contains \(\displaystyle \sqrt{-x}\) in which case the limit could exist at negative values of x but won't exist at positive values of x! You should make sure you examine what is actually under the radical.