Can someone tell how to answer this problem using indeterminate form?

http://i104.photobucket.com/albums/m...emike1/ind.jpg

Thanks in advance..

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- Oct 8th 2010, 02:34 AMcutiemike1Indeterminate Forms
Can someone tell how to answer this problem using indeterminate form?

http://i104.photobucket.com/albums/m...emike1/ind.jpg

Thanks in advance.. - Oct 8th 2010, 04:20 AMArchie Meade
Logs can help in such cases...

$\displaystyle L=\displaystyle\lim_{x \to 0}\left(1+x^2\right)^{\frac{4}{x}}$

$\displaystyle \displaystyle\ log_eL=\lim_{x \to 0}log_e\left(1+x^2\right)^{\frac{4}{x}}=\lim_{x \to 0}\left[\frac{log_e\left(1+x^2\right)^4}{x}\right]$

and using L'Hopital's rule, since we have a 0/0 indeterminate form

$\displaystyle \displaystyle\ =\lim_{x \to 0}\frac{4\left(1+x^2\right)^32x}{\left(1+x^2\right )^4}$

$\displaystyle \displaystyle\ =\lim_{x \to 0}\frac{8x\left(1+x^2\right)^3}{\left(1+x^2\right) ^3\left(1+x^2\right)}$

$\displaystyle \displaystyle\lim_{x \to 0}\frac{8x}{1+x^2}=0$

$\displaystyle log_eL=0\Rightarrow\ e^0=L=1$ - Oct 8th 2010, 04:48 AMHallsofIvy
There is no reason not to take that "4" out of the exponent along with the "1/x":

$\displaystyle ln(L)= 4\lim_{x\to 0}\frac{ln(1+ x^2)}{x}$

and, by L'Hopital's rule

$\displaystyle ln(L)= 4\lim_{x\to 0}\frac{2x}{1+ x^2}$

which is 0. - Oct 8th 2010, 04:53 AMArchie Meade
- Oct 8th 2010, 04:56 AMcutiemike1
I thought of that one but on my part I used natural logarithm. When I was solving it using natural logarithm I suddenly didn't know what must be done. This one will help me a lot in my final exam.

**MANY THANKS TO YOU!!**

Now my questions are:

1. What might be the other cases which uses Logs?

2. Is this always true whenever a limit of a polynomial which is raised to a fraction whose denominator approaches to zero?

3. Natural Logarithms can also be?

4. And what if for example a limit approaches to either POSITIVE of NEGATIVE Infinity, how could it be substituted to the function given, what would be the answer and how should it be done?

THANKS AGAIN IN ADVANCE!!! - Oct 8th 2010, 04:59 AMcutiemike1
- Oct 8th 2010, 05:03 AMcutiemike1
- Oct 8th 2010, 06:33 AMArchie Meade
- Oct 8th 2010, 06:45 AMArchie Meade
- Oct 8th 2010, 03:31 PMcutiemike1
I understood the whole thing now. Thanks a lot to the both of you.