# Undefined limit?

1. Dec 23, 2007

### uman

For the scantron-based (*sigh*) final of my high school calculus class, one of the problems was to evaluate
$$\lim_{x\to 4^+}\sqrt{4-x}$$.

I answered "undefined", reasoning that since $$\sqrt{4-x}$$ is undefined for all x "to the right of" 4, it could hardly approach any value as x approaches 4 from the right.

More formally, if the limit exists then there is some number L such that for any strictly postive number $$\epsilon$$, there is a strictly postive number $$\delta$$ such that
$$|\sqrt{4-x}-L|<\epsilon$$ whenever $$4<x<4+\delta$$. Since x is always greater than 4, $$|\sqrt{4-x} - L|$$ is undefined (because negative numbers are not in the domain of the square root function) and therefore not less than $$\epsilon$$. This leads to a contradiction, therefore there is no limit as x approaches 4 from the right.

My teacher disagreed and said that if the scantron marked me wrong, I must be wrong. She had very little time to listen to me as she had to go eat lunch or something.

Is there anything wrong with my reasoning, or am I right? My final grade in the class could depend on this, so if I'm not mistaken somewhere I plan on confronting my teacher about it after winter break.

Thank you.

Last edited: Dec 23, 2007
2. Dec 23, 2007

### EnumaElish

Since the function is defined for x < 4 and undefined for x > 4, the right-limit is not applicable. "The" limit is defined as the left-limit, which is the only applicable direction here.

Put differently, the only admissible sequence approaching from the right is the constant sequence c = {4, 4, ...}. Since f(c) = 0, the right-limit condition is satisfied in a trivial way.

Last edited: Dec 23, 2007
3. Dec 23, 2007

### uman

So the right-hand limit is undefined, yes?

4. Dec 23, 2007

### EnumaElish

The concept of a right limit does not apply to this example; the only directional limit that is applicable is the left-limit, which is well-defined. Therefore "the" limit exists and is well defined.

I am sorry for your grades; hopefully they'll be better next semester.

5. Dec 23, 2007

### uman

Yes, "the" limit exists and is equal to the left-hand limit. That's not what the problem asks. The problem explicity asked for the right-hand limit, not the left-hand limit or the limit in general. The right-hand limit is, as you say, meaningless here and therefore imo "undefined" is the best answer.

Thanks for your sympathy.

6. Dec 23, 2007

### LukeD

uman. If it does indeed ask for the right hand limit here, then your answer is correct. The function is not defined on that interval, so the concept for the right hand limit does not make any sense. (It would be like asking for the largest element of the empty set. The request makes no sense.)

The only way your answer would be wrong would be if you were allowing the square root function to be defined for negative values. (Though from your discussion, I'm assuming that you've defined it as being a function from non-negative real numbers to non-negative real numbers). In this case, though the expression would have an imaginary value for all x>4, the limit would exist and be equal to 0.

So really, whether or not you got the question right depends on if it was really asking for the right hand limit (perhaps you remembered the question incorrectly) and if you are allowing negative numbers to have square roots.

Last edited: Dec 23, 2007
7. Dec 23, 2007

### uman

Thanks for the response.

Since we haven't learned about complex numbers, I can't see how the sqrt of a negative number would be defined. And I'm almost positive I didn't read it wrong.

I'm going to challenge my grade then, thanks.

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