Find the value of the one sided limit
$\displaystyle \lim_{x \to 2} \frac{4-x^2}{\sqrt{2+x-x^2}}$
WHere will the lim be shifting from the left( - ) or from the right ( + )? and why?
Thank you very much for all your help!
Since both the numerator and the denominator $\displaystyle \longrightarrow 0$, use l’Hopital’s rule.
Factor, your idea was correct.
$\displaystyle \lim_{x\to 2} \frac{(2-x)(2+x)}{\sqrt{(2-x)(1+x)}}$
Now when we approach the limit from the right,
$\displaystyle x>2$ that means,
$\displaystyle 2-x<0$
$\displaystyle 1+x>0$
But then,
$\displaystyle (2-x)(1+x)<0$
And the function is undefined.
(Radical is a negative number!)
So the limit (the regular limit) does not exist.
However, the one-handed limit, might. We already shown that the right-handed does not exists. Let us see the left-handed.
When we approach the limit from the left,
$\displaystyle x<2$ on some open interval $\displaystyle (\delta, 2)$ where $\displaystyle \delta <2$ (a really small number). Because we do not care what happens far away one what happens really close to the point.
In that case,
$\displaystyle 2-x>0$
$\displaystyle 1+x>0$
Thus, the radical is defined.
And furthermore we can factor,
$\displaystyle \lim_{x\to 2^-} \frac{(2-x)(2+x)}{\sqrt{2-x}\cdot \sqrt{1+x}}$
$\displaystyle \lim_{x\to 2^-} \frac{\sqrt{2-x}(2+x)}{\sqrt{1+x}}=\frac{0\cdot 4}{3}=0$
I don't have an example of where it doesn't work, I was simply speculating. I was just trying to be careful since the derivative of $\displaystyle \sqrt{2 + x - x^2}$ doesn't exist at x = 2 (because the function doesn't exist for x < 2.) Apparently as long as the limit of the derivative exists, L'Hopital is usable, if I'm understanding TPH correctly.
-Dan
First I would like to mention square roots of of negative numbers are undefined. Note, it is mathematically incorrect to say $\displaystyle \sqrt{-4}=2i$ (I know that is what they teach in high schools, but it is still wrong). I have already discussed that on the forum. Next, all the L'Hopitals rule was only proven for real differenciable functions (at least I think). Finally this problem it self is real, implies that the function the poster was using mapped the real numbers into the real numbers.