1
$\begingroup$

When I'm trying to find the limit of $\frac{\sqrt{1-\cos x}}{\sin x}$ when x approaches 0, using power series with "epsilon function" notation, it goes :

$\dfrac{\sqrt{1-\cos x}}{\sin x} = \dfrac{\sqrt{\frac{x^2}{2}+x^2\epsilon_1(x)}}{x+x\epsilon_2(x)} = \dfrac{\sqrt{x^2(1+2\epsilon_1(x))}}{\sqrt{2}x(1+\epsilon_2(x))} = \dfrac{|x|}{\sqrt{2}x}\dfrac{\sqrt{1+2\epsilon_1(x)}}{1+\epsilon_2(x)} $

But I can't seem to do it properly using Landau notation

I wrote :

$ \dfrac{\sqrt{\frac{x^2}{2}+o(x^2)}}{x+o(x)} $

and I'm stuck... I don't know how to carry these o(x) to the end

Could anyone please show me what the step-by-step solution using Landau notation looks like when written properly ?

$\endgroup$
3
  • $\begingroup$ With Landau, $o(x)=xo(1)$ and $o(x^2)=x^2o(1)$. And $o(1)$ just means the same as $\epsilon(x)\rightarrow 0$. $\endgroup$
    – Julien
    May 16 '13 at 13:56
  • $\begingroup$ This limit doesn't exist unless you have $x\to 0+$. When $x\to 0-$ the value is different. $\endgroup$ May 16 '13 at 14:03
  • $\begingroup$ Ok, i got it. Thanks $\endgroup$ May 16 '13 at 14:03
2
$\begingroup$

It is the same as in the "$\epsilon$" notation. For numerator, we want $\sqrt{x^2\left(\frac{1}{2}+o(1)\right)}$, which is $|x|\sqrt{\frac{1}{2}+o(1)}$. In the denominator, we have $x(1+o(1))$.

Remark: Note that the limit as $x\to 0$ does not exist, though the limit as $x$ approaches $0$ from the left does, and the limit as $x$ approaches $0$ from the right does.

$\endgroup$
1
$\begingroup$

If you are willing to skip Landau notation, you can just note (when $0<x<\pi$), $\sin x = \sqrt{\sin^2 x}=\sqrt{1-\cos^2 x}$. Then:

$$\frac{\sqrt{1-\cos x}}{\sin x} = \sqrt{\frac{1-\cos x}{1-\cos^2 x}}=\frac{1}{\sqrt{1+\cos x}}$$

Note this is why there is no general limit as $x\to 0$. When $-\pi<x<0$, the value of the expression is $\frac{-1}{\sqrt{1+\cos x}}$. You can only get a limit as $x\to 0+$ or as $x\to 0-$, not a general limit.

$\endgroup$
0
$\begingroup$

Now $$\frac{\sqrt{1 - \cos(x)}}{\sin(x)} = \frac{\sqrt{2 \sin^2(x/2)}}{\sin(x)} = \sqrt{2} \frac{ |\sin(x/2)|}{\sin(x)} = \sqrt{2} \frac{|x/2| + O(x^3)}{x + O(x^3)}\\ = \sqrt{2} \frac{\text{sgn}(x)/2 + O(x^2)}{1 + O(x^2)} = \text{sgn}(x)/\sqrt{2} + O(x^2)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.