# Today test - one limit

1. Jan 7, 2005

### twoflower

Hi all,

could you help me with the limit we had in today test please? Here it is:

Find out, for which $\alpha \in \mathbb{R}$ the limit exists and is finite:

$$\lim_{x \rightarrow 0_{+}} \frac{ \log \left( \sqrt{1 + x^2} - x \right) - \sin \left( x^{\alpha} \right)}{x^{\alpha}}$$

and find the limit for those $\alpha$

I tried it, but I had too many questionable steps there, so I won't post it here till I find out what is the right solution

Thank you.

2. Jan 7, 2005

### dextercioby

Let's take the most simple case,viz.$\alpha =0$

Can u compute the limit in that case??

Daniel.

PS.Provide the calculations and the result.

3. Jan 7, 2005

### twoflower

Ok, here is how I did it:

$$\lim_{x \rightarrow 0_{+}} \frac{ \log \left( \sqrt{1 + x^2} - x \right) - \sin \left( x^{\alpha} \right)}{x^{\alpha}} =$$

$$\lim_{x \rightarrow 0_{+}} \frac{ \log \left( 1 + \left(\sqrt{1 + x^2} - x - 1 \right) \right) }{x^{\alpha} \left( \sqrt{1 + x^2} - x - 1 \right)} . \lim_{x \rightarrow 0_{+}} \left( \sqrt{1 + x^2} - x - 1 \right) - \lim_{x \rightarrow 0_{+}} \frac{\sin \left( x^{\alpha} \right)}{x^{\alpha}} =$$

$$0.\lim_{x \rightarrow 0_{+}} \frac{1}{x^{\alpha}} - \lim_{x \rightarrow 0_{+}} \frac{\sin \left( x^{\alpha} \right)}{x^{\alpha}}$$

In this place I thought that $\alpha$ must be < 0 so that the limit will be finite (because of the $\frac{1}{x^{\alpha}}$ limit). Then my steps are ok and in result the limit is 0

4. Jan 7, 2005

### dextercioby

$$L=\lim_{x \rightarrow 0_{+}} \frac{ \ln \left( \sqrt{1 + x^2} - x \right) - \sin \left( x^{\alpha} \right)}{x^{\alpha}}$$

Case a)$\alpha =0$
Then
$$x^{\alpha}=1;\sin(x^{\alpha})=\sin 1$$

The limit becomes:
$$L=\lim_{x\searrow 0} \ln(\sqrt{1+x^{2}}-x) -\sin 1=0-\sin 1=-\sin 1$$ (1)

Case b)$\alpha >0$
Then u can apply August L'Ho^spital rule to get:

$$\lim_{x\searrow 0} \frac{x-\sqrt{1+x^{2}}}{\alpha\sqrt{1+x^{2}}(\sqrt{1+x^{2}}-x) x^{\alpha -1}}-1$$ (2)

U have 3 cases:

b1)$\alpha>1$
The limit is:
$$L=-\infty$$ (3)

b2)$\alpha=1$
The limit is:
[tex] L=-1-1=-2 [/itex] (4)

b3)$0< \alpha <1$
The limit is:
[tex] L=0-1=-1 [/itex] (5)

I let u now consider the case "c)",when $\alpha < 0$.

Daniel.