Indeterminate Forms

• Oct 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

• Oct 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...

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

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

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 AM
HallsofIvy
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 AM
Quote:

Originally Posted by HallsofIvy
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.

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

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?

• Oct 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":

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

Sorry HallsofIvy, I didn't notice this reply. I am currently acknowledging the help of Archie Meade.
So the final answer is 0?
• Oct 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":

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

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

$\displaystyle ln(L)=\log_e(L)=0\Rightarrow\ L=e^0=1$
• Oct 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.... $\displaystyle lnx=\log_ex$

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?