# Indeterminate Forms

• October 8th 2010, 02:34 AM
cutiemike1
Indeterminate Forms
Can someone tell how to answer this problem using indeterminate form?

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

• October 8th 2010, 04:20 AM
Quote:

Originally Posted by cutiemike1
Can someone tell how to answer this problem using indeterminate form?

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

Logs can help in such cases...

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

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

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

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

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

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

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

$ln(L)= 4\lim_{x\to 0}\frac{ln(1+ x^2)}{x}$
and, by L'Hopital's rule
$ln(L)= 4\lim_{x\to 0}\frac{2x}{1+ x^2}$
which is 0.
• October 8th 2010, 04:53 AM
Quote:

Originally Posted by HallsofIvy
There is no reason not to take that "4" out of the exponent along with the "1/x":

$ln(L)= 4\lim_{x\to 0}\frac{ln(1+ x^2)}{x}$
and, by L'Hopital's rule
$ln(L)= 4\lim_{x\to 0}\frac{2x}{1+ x^2}$
which is 0.

Thanks HallsofIvy.
That's a huge improvement!
• October 8th 2010, 04:56 AM
cutiemike1
Quote:

Originally Posted by Archie Meade
Logs can help in such cases...

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!!!
• October 8th 2010, 04:59 AM
cutiemike1
Quote:

Originally Posted by HallsofIvy
There is no reason not to take that "4" out of the exponent along with the "1/x":

$ln(L)= 4\lim_{x\to 0}\frac{ln(1+ x^2)}{x}$
and, by L'Hopital's rule
$ln(L)= 4\lim_{x\to 0}\frac{2x}{1+ x^2}$
which is 0.

Sorry HallsofIvy, I didn't notice this reply. I am currently acknowledging the help of Archie Meade.
So the final answer is 0?
• October 8th 2010, 05:03 AM
cutiemike1
Quote:

Originally Posted by HallsofIvy
There is no reason not to take that "4" out of the exponent along with the "1/x":

$ln(L)= 4\lim_{x\to 0}\frac{ln(1+ x^2)}{x}$
and, by L'Hopital's rule
$ln(L)= 4\lim_{x\to 0}\frac{2x}{1+ x^2}$
which is 0.

Sorry HallsofIvy, I didn't notice this reply. I am currently acknowledging the help of Archie Meade.
So the final answer is 0?
• October 8th 2010, 06:33 AM
Quote:

Originally Posted by cutiemike1
Sorry HallsofIvy, I didn't notice this reply. I am currently acknowledging the help of Archie Meade.
So the final answer is 0?

Hi cutiemike1,

The natural logarithm is the logarithm to the base "e".
HallsofIvy showed a better way to reach ln(L).
Finally, we evaluate L.

$ln(L)=\log_e(L)=0\Rightarrow\ L=e^0=1$
• October 8th 2010, 06:45 AM
Quote:

Originally Posted by cutiemike1
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? typically in cases where the variable is in the exponent

Logarithm - Wikipedia, the free encyclopedia

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

L'Hopital's rule is used in cases where both numerator and denominator
of a quotient are both approaching 0 or both approaching infinity.

To apply that rule, both numerator and denominator are differentiated.

3. Natural Logarithms can also be? The natural logarithm is defined as.... $lnx=\log_ex$

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!!!

I don't have a suitable answer to (4.) above.
Hopefully you will receive a reply, but I'd advise maybe starting a new thread,
quoting some specific examples and how far you can get with them.
You'd get some good responses then as logs are not my forte!
• October 8th 2010, 03:31 PM
cutiemike1
I understood the whole thing now. Thanks a lot to the both of you.