What is the Limit for Today's Test on Logarithms and Trigonometry?

[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:

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

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.

 

[dextercioby]

Let's take it gradually:
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.

 

[twoflower]

Ok, here is how I did it:

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

<br /> \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}} = <br />

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

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

 

[dextercioby]

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

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 &gt;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&gt;1
The limit is:
L=-\infty (3)

b2)\alpha=1
The limit is:
L=-1-1=-2 [/itex] (4)<br /> <br /> b3)0&amp;lt; \alpha &amp;lt;1<br /> The limit is:<br /> L=0-1=-1 [/itex] (5)&lt;br /&gt; &lt;br /&gt; I let u now consider the case &amp;quot;c)&amp;quot;,when \alpha &amp;amp;lt; 0.&lt;br /&gt; &lt;br /&gt; Daniel.