Limits of a logarithm

Bud · #1 $Old$ December 3rd, 2008, 01:25 AM

Hi folks,

I have some problems with this limit:

I cannot decide which direction to go. The 1+... can be neglected right? But what about the constants in the exponent?

Thanks in advance.
Bud

mr fantastic · #2 $Old$ December 3rd, 2008, 01:43 AM

Quote:

Originally Posted by Bud $View Post$

Hi folks,

I have some problems with this limit:

I cannot decide which direction to go. The 1+... can be neglected right? Mr F says: Wrong.

But what about the constants in the exponent?

Thanks in advance.
Bud

Start by considering $\lim_{z \rightarrow + \infty} \left[ 1 + C e^{A(z-B)}\right]$ :

C > 0:

A > 0: The limit is equal to +oo.
A < 0: The limit is equal to 1.

C < 0:

A < 0: The limit is equal to 1.
A > 0: The limit is equal to -oo.

Now consider $\ln \lim_{z \rightarrow + \infty} \left[ 1 + C e^{A(z-B)}\right]$ (why is this change of order allowed?) ..... (One of the abvove cases is not valid).

Bud · #3 $Old$ December 3rd, 2008, 01:59 AM

Quote:

Originally Posted by mr fantastic $View Post$

Start by considering $\lim_{z \rightarrow + \infty} \left[ 1 + C e^{A(z-B)}\right]$ :

C > 0:

A > 0: The limit is equal to +oo.
A < 0: The limit is equal to 1.

C < 0:

A < 0: The limit is equal to 1.
A > 0: The limit is equal to -oo.

Now consider $\ln \lim_{z \rightarrow + \infty} \left[ 1 + C e^{A(z-B)}\right]$ (why is this change of order allowed?) ..... (One of the abvove cases is not valid).

Hi,

we can stick to the case C>0, A>0. I have a physical problem here and I know that the constants are always positive. This leaves us with: The limit is equal to +oo.

I do not know why the change of order is allowed. The case for which the change is not valid is the case when the limit goes to -oo. Here the logarithm is not defined.