# Thread: what is the limit of the natural log

1. ## what is the limit of the natural log

find the limit of the seq:

ln ( 2 x ^ 2 + 1) - ln ( x ^ 2 + 1)

lim ln [( 2 x ^2 +1)/(x ^ 2 + 1)]
x->inf

= ln(3/2) and if we say product of logs then the solution to the limit is:

ln(3)-ln(2)

Problem is that when I drop the initial problem into my calculator I get an answer of log(2).

which is right?

2. Originally Posted by orendacl
find the limit of the seq:

ln ( 2 x ^ 2 + 1) - ln ( x ^ 2 + 1)

lim ln [( 2 x ^2 +1)/(x ^ 2 + 1)]
x->inf

= ln(3/2) and if we say product of logs then the solution to the limit is:

ln(3)-ln(2)

Problem is that when I drop the initial problem into my calculator I get an answer of log(2).

which is right?
lim [( 2 x ^2 +1)/(x ^ 2 + 1)] = 2
x->inf

So
lim ln [( 2 x ^2 +1)/(x ^ 2 + 1)] = ln 2
x->inf

(I'm not sure where you got 3/2 from???)

3. Originally Posted by orendacl
find the limit of the seq:

ln ( 2 x ^ 2 + 1) - ln ( x ^ 2 + 1)

lim ln [( 2 x ^2 +1)/(x ^ 2 + 1)]
x->inf

= ln(3/2) and if we say product of logs then the solution to the limit is:

ln(3)-ln(2)

Problem is that when I drop the initial problem into my calculator I get an answer of log(2).
Well of course you got $\displaystyle \log(2)$ that is the correct answer!
$\displaystyle \left(\frac{2x^2+1}{x^2+1}\right)\to 2$.

4. Originally Posted by Plato
Well of course you got $\displaystyle \log(2)$ that is the correct answer!
$\displaystyle \left(\frac{2x^2+1}{x^2+1}\right)\to 2$.
The answer is ln 2 NOT log 2

5. Originally Posted by Debsta
The answer is ln 2 NOT log 2
Professional mathematicians use $\displaystyle \log(x)$ where others use $\displaystyle \ln(x)$.

6. Originally Posted by Plato
Professional mathematicians use $\displaystyle \log(x)$ where others use $\displaystyle \ln(x)$.
I don't think orendacl is a "professional mathematician".

7. log 2 = .6931472
ln 2 = .30103

These are not the same...

8. Originally Posted by Debsta
I don't think orendacl is a "professional mathematician".
Originally Posted by orendacl
Problem is that when I drop the initial problem into my calculator I get an answer of log(2).
Did you notice that orendacl reported that the calculator gave the answer as $\displaystyle \log(2)$?
Orendacl seems to have a calculator meant for professional use.

For those who are so far behind as to use a base 10 logarithm, write it as $\displaystyle \log_{10}(x)$.
And BTW note the usage of the (). We are using function notation.

9. Thanks for the reply plato. Your analysis is indeed correct. I have several different math tools or techniques to solve and or check a problem and though not a profi myself, use the tools each one when a problem persists, and as an aid to see which is most reliable.

Much appreciated to you all!

10. Originally Posted by Plato
Did you notice that orendacl reported that the calculator gave the answer as $\displaystyle \log(2)$?
Orendacl seems to have a calculator meant for professional use.

For those who are so far behind as to use a base 10 logarithm, write it as $\displaystyle \log_{10}(x)$.
And BTW note the usage of the (). We are using function notation.
I don't know why some mathematicians adopt such an arrogant stand. If I am one of those you (Plato) call "so far behind" I am greatly offended. The reason why I help on this site is that I see younger students struggling with what you could call rather basic concepts. But for them it is so frustrating when people help in language they do not understand .... yet. I always try to pitch my advice at a level they are likely to understand, while also trying to keep the mathematics correct. At that level the distinction between ln and log causes great confusion. The use of () adds further confusion. Yes I understand what you are saying, but I don't think those subtleties are important at the stage of the question I was addressing.

11. Originally Posted by orendacl
Thanks for the reply plato. Your analysis is indeed correct. I have several different math tools or techniques to solve and or check a problem and though not a profi myself, use the tools each one when a problem persists, and as an aid to see which is most reliable.
Originally Posted by Debsta
I don't know why some mathematicians adopt such an arrogant stand. If I am one of those you (Plato) call "so far behind" I am greatly offended. The reason why I help on this site is that I see younger students struggling with what you could call rather basic concepts. But for them it is so frustrating when people help in language they do not understand .... yet. I always try to pitch my advice at a level they are likely to understand, while also trying to keep the mathematics correct. At that level the distinction between ln and log causes great confusion. The use of () adds further confusion. Yes I understand what you are saying, but I don't think those subtleties are important at the stage of the question I was addressing.
Did you bother to read the first quote above?
You say that “The use of () adds further confusion” which means that you are contributing to the ignorance of function notation. Why would anyone do that?
If you indeed want to want to help younger students then you owe it to them to be up to date on current practices such as a calculator giving the answer $\displaystyle \log(2)$.

12. Originally Posted by Plato
Did you bother to read the first quote above?
You say that “The use of () adds further confusion” which means that you are contributing to the ignorance of function notation. Why would anyone do that?
If you indeed want to want to help younger students then you owe it to them to be up to date on current practices such as a calculator giving the answer $\displaystyle \log(2)$.
I don't think I am "contributing to the ignorance of function notation". Could you please explain the practical difference to me between a result of log 2 and log(2)?