Need help discussing a limit.

[wewfy]

Hello everybody, im stuck with this discussion, do you have any suggestions?

[MATH]\displaystyle \lim{x \to 0} \hspace{0,25in} \begin{equation*}\frac{Ln(1+tan(x))-Ln(1+sin(x))}{x^\alpha} \end{equation*}[/MATH]

 

[Dr.Peterson]

Have you tried L'Hopital's rule? If so, what happened?

If you haven't learned that yet, what have you learned that might be relevant?

In other words, show us where you're stuck, so we can try to pull you out.

 

[wewfy]

Im sorry, I should have written that I can only use equivalents, and those x's should be xn of successions..

 

[Dr.Peterson]

Im sorry, I should have written that I can only use equivalents, and those x's should be xn of successions..

I'm not sure what you're saying the problem really is. If the x's should be a sequence \(x_n\), then you would be taking a limit as n approaches infinity, not as x approaches zero.

In addition, I forgot to ask what \(\alpha\) is (the exponent); its value can make a huge difference.

Please give us an image of the actual complete problem (even if it is not in English). And if you were "stuck" without actually trying anything (as I have to guess, because you still show no work), then maybe you can show us an example you were given so we can see what is allowed.

 

[wewfy]

I have to discuss the limit for [MATH]α \in \mathbb{R} [/MATH][MATH]Xn \rightarrow 0[/MATH]
Here is my work, I think I made a mistake assuming that [MATH]\frac{1}{1+sin(x_n)} \rightarrow 1[/MATH]

 

[Dr.Peterson]

Your method of "equivalents" is somewhat equivalent to L'Hopital's rule. I didn't know that was what you meant, and am not sure exactly what is allowed. The step you ask about is clearly true, simply because the sine is continuous.

But it appears that your final step will be to determine for what values of alpha the limit exists, and what it is in each case. What thoughts do you have there?

 

[wewfy]

When alpha is three the limit would be 1/2

When alpha < 3, converges to 0

And when alpha > 3, diverges to +inf