Limits dealing with infinity

[tjkubo]

I've been learning about the derivative of natural logarithms in order to define e. So far we got this equation by applying the definition of the derivative:

d/dx ln x = ln lim n→0 (1 + x/h)^1/h

After chaning the variable to n,

d/dx ln x = ln lim n→∞ (1 + 1/nx)^n

This is my question: Can't you simplify the limit here?
Since lim n→∞ (1/n) = 0 and 1^∞ = 1

lim n→∞ (1 + 1/nx)^n = (1 + 0)^∞ = 1

Therefore,
d/dx ln x = ln 1 = 0

I know I'm probably wrong, but what mistake am I'm making? Is is just conceptual?

 

[tkhunny]

\(\displaystyle (1+1/1)^{1} = 2\)
\(\displaystyle (1+1/2)^{2} = 2.25\)
\(\displaystyle (1+1/4)^{4} = 2.44140625\)
\(\displaystyle (1+1/8)^{8} = 2.565784514\)
\(\displaystyle (1+1/16)^{16} = 2.637928497\)
\(\displaystyle (1+1/100)^{100} = 2.704813829\)
\(\displaystyle (1+1/1000)^{1000} = 2.716923932\)
\(\displaystyle (1+1/10000)^{10000} = 2.718145927\)

That is monotonic increasing. You tell me if it was OK to simplify that limit. Is your simplification consistent with this result?

 

[galactus]

Using the derivative definition:

\(\displaystyle \L\\\frac{d}{dx}[ln(x)]=\lim_{h\to\0}\frac{ln(x+h)-ln(x)}{h}\)

\(\displaystyle \L\\=\lim_{h\to\0}\frac{1}{h}ln(\frac{x+h}{h})\)

\(\displaystyle \L\\=\lim_{h\to\0}\frac{1}{h}ln(1+\frac{h}{x})\)

Now, let \(\displaystyle n=\frac{h}{x},\;\ n\to\0\;\ as\;\ h\to\0\)

\(\displaystyle \L\\\lim_{n\to\0}\frac{1}{nx}ln(1+n)\)

\(\displaystyle \frac{1}{x}\) doesn't vary with n, so it can be moved through the limit sign:

\(\displaystyle \L\\=\frac{1}{x}\lim_{n\to\0}\frac{1}{n}ln(1+n)\)

Property of logs:

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

Because of continuity we can do this:

\(\displaystyle \L\\=\frac{1}{x}ln[\overbrace{\lim_{n\to\0}(1+n)^{\frac{1}{n}}}^{e}]\)

\(\displaystyle \L\\=\frac{1}{x}ln(e)=\H\\\frac{1}{x}\)

 

[tjkubo]

Thanks for the replies.

I always get confused when it comes to e.
How do you know lim n→0 (1 + n)^1/n = e
or that lim n→∞ (1 + 1/n)^n = e?

Is there a graphical solution to that?

 

[tkhunny]

"Graphical solution"? What is that? There are graphical hints. See my post, above. That never will constitute proof or a "solution". This is mathematics, not art.

 

[galactus]

\(\displaystyle y=(1+\frac{1}{n})^{n}\) has a horizontal asymptote at e. I believe it was Euler who discovered that.

TKH's chart shows you.


The proof of this is built around the differentiability of ln(x), at 1 in particular.

\(\displaystyle 1=\lim_{n\to\0}\frac{ln(1+n)-ln(1)}{n}=\lim_{n\to\0}\frac{ln(1+n)}{n}=\lim_{n\to\0}ln(1+n)^{\frac{1}{n}}\)

Therefore and whence,

\(\displaystyle e=e^{\lim_{n\to\0}ln(1+h)^{\frac{1}{n}}}\)

Remember the continuous thing:

\(\displaystyle e=\lim_{n\to\0}e^{ln(1+n)^{\frac{1}{n}}}\)

\(\displaystyle \L\\=\lim_{n\to\0}(1+n)^{\frac{1}{n}}\)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

\(\displaystyle \L\\\lim_{n\to\infty}(1+\frac{1}{n})^{n}=e\) can be derived

from \(\displaystyle \L\\\lim_{n\to\0}(1+n)^{\frac{1}{n}}=e\) by making

the correct substitutions.