Finding the Limit: One-sided and Imaginary Numbers

[Kilgour22]

Hi all! In the midst of making up limit problems to solve generally, I came across a limit in which I'm not sure of the answer. This is not homework, but merely a product of free time.

Homework Statement

Find the limit.​

$lim_{x \to a^+} {\frac{\sqrt(a^2 - x^2)}{a}} , a > 0$​

The attempt at a solution

$lim_{x \to a^+} {\frac{\sqrt(a^2 - x^2)}{a}}$

$= lim_{x \to a^+} {\frac{a\sqrt(1 - (x/a)^2)}{a}}$

$= lim_{x \to a^+} {\sqrt(1 - (x/a)^2)}$

$= \sqrt{1 - ((a + b)/a)^2}, b > 0$

$= \sqrt{1 - ((a^2 + 2ab + b^2)/a^2)}$

$= \sqrt{1 - (1 + 2b/a + (b/a)^2)}$

$= \sqrt{-2b/a - (b/a)^2} = i\sqrt{2b/a + (b/a)^2}, a >0, b > 0$​

Because I'm left with an imaginary answer, does that mean that the one-sided limit does not exist, or would the above expression actually be the answer? I've never had to deal with limits that yield imaginary numbers before, so any clarification as to how they are handled would be much appreciated!

 

[Ray Vickson]

Hi all! In the midst of making up limit problems to solve generally, I came across a limit in which I'm not sure of the answer. This is not homework, but merely a product of free time.

Homework Statement

Find the limit.​

$lim_{x \to a^+} {\frac{\sqrt(a^2 - x^2)}{a}} , a > 0$​

The attempt at a solution

$lim_{x \to a^+} {\frac{\sqrt(a^2 - x^2)}{a}}$

$= lim_{x \to a^+} {\frac{a\sqrt(1 - (x/a)^2)}{a}}$

$= lim_{x \to a^+} {\sqrt(1 - (x/a)^2)}$

$= \sqrt{1 - ((a + b)/a)^2}, b > 0$

$= \sqrt{1 - ((a^2 + 2ab + b^2)/a^2)}$

$= \sqrt{1 - (1 + 2b/a + (b/a)^2)}$

$= \sqrt{-2b/a - (b/a)^2} = i\sqrt{2b/a + (b/a)^2}, a >0, b > 0$​

Because I'm left with an imaginary answer, does that mean that the one-sided limit does not exist, or would the above expression actually be the answer? I've never had to deal with limits that yield imaginary numbers before, so any clarification as to how they are handled would be much appreciated!

Your original (pre-limit) expression contains only $x$ and $a$, so how on Earth could you possibly be getting an answer involving some $b$ that was not part of the problem at all? Anyway, if $x > a$ is near $a$, we can write $x = a + h$, where $h > 0$ is a small quantity. Now re-write your ratio $f(x) = \frac{1}{a} \sqrt{a^2 - x^2}$ in terms of $a$ and $h$. That should reveal much more clearly what is going on when you take $h \to 0+$.

 

[Kilgour22]

Your original (pre-limit) expression contains only $x$ and $a$, so how on Earth could you possibly be getting an answer involving some $b$ that was not part of the problem at all? Anyway, if $x > a$ is near $a$, we can write $x = a + h$, where $h > 0$ is a small quantity. Now re-write your ratio $f(x) = \frac{1}{a} \sqrt{a^2 - x^2}$ in terms of $a$ and $h$. That should reveal much more clearly what is going on when you take $h \to 0+$.

I let $x = a + b$. I probably should have mentioned that, though I thought I made the steps pretty clear. If I let $b \to 0^+$ , I get the limit to be 0. So I guess that's the answer, even though it's imaginary up until that point?

 

[HallsofIvy]

The very first thing I notice is that $\sqrt{a^2- x^2}$ is NOT a real number for x> a so that $\lim_{x\to a^+}$ will not exist! Are you sure you have written the problem correctly?

If you meant $\lim_{x\to a^-}\frac{\sqrt{a^2-x^2}}{a}$ then you should see immediately that the numerator is going to 0 while the denominator is not. I see no reason to do any of the calculations you have done.

 

[Kilgour22]

Thank you, that's precisely what I needed to know!