Once again, many thanks for your reply Lex. I feel like I've gone down the rabbit hole- one answer leads to many more questions. I suppose this is the nature of inquiry. Is there a way I can save this thread as a PDF for future reference?This was my proof for (c*):

[imath]\log_a x[/imath] is continuous at [imath]x=k>0[/imath]

so [imath]\log_a \lim\limits_{u \to k} u = \lim\limits_{u \to k} \log_a u[/imath]

[imath]\therefore \log_a \lim \limits_{x \to \infty}\left(1+\tfrac{1}{x}\right)^x = \lim \limits_{x \to \infty} \log_a \left(1+\tfrac{1}{x}\right)^x[/imath]

Expanding on this.

To be able to 'reverse the order of the function and the limit', we want the limit to exist and the function to be continuous at the value of the limit.

You can use the ready-made theorem:

View attachment 28321

In the present case, [imath]f(X)=\log X[/imath] and [imath]g(x)=\left(1+\tfrac{1}{x}\right)^x[/imath]

Unfortunately [imath]\log X[/imath] is not continuous at X=0, so we have to show that [imath]\lim \limits_{x \to \infty}\left(1+\tfrac{1}{x}\right)^x[/imath] exists and [imath]=k>0[/imath].

[imath]\left(1+\tfrac{1}{x}\right)^x >1[/imath] for all x>0, therefore if the limit exists, [imath]\lim\limits_{x \to \infty} \left(1+\tfrac{1}{x}\right)^x=k≥1>0[/imath]

The only remaining problem is to prove that the limit exists.

It is sufficient to prove that [imath]\left(1+\tfrac{1}{x}\right)^x[/imath] is increasing, and bounded above by [imath]e[/imath] as there is a theorem which says that an increasing function, bounded above, converges to a limit.

[imath]X>1[/imath]

[imath]\ln X+\dfrac{1}{X}-1[/imath] is strictly increasing and therefore [imath]>0[/imath], when [imath]X>1\hspace3ex (\text{the derivative }\tfrac{1}{X}-\tfrac{1}{X^2}>0, X>1 \text{ and }\ln 1+\tfrac{1}{1}-1=0[/imath])

[imath]\therefore \ln(\tfrac{x+1}{x})+\tfrac{x}{x+1}-1 >0[/imath] (decreasing), x>0 (since [imath]X=\tfrac{x+1}{x} >1[/imath], (decreasing), [imath]x>0[/imath])

i.e. [imath]\ln(1+\tfrac{1}{x})-\tfrac{1}{1+x}>0[/imath], (decreasing), [imath]x>0[/imath]

i.e. [imath]\dfrac{\text{d}}{\text{dx}}[x\ln(1+\tfrac{1}{x})]>0[/imath], (decreasing), [imath]x>0[/imath]

[imath]\therefore \ln (1+\tfrac{1}{x})^x[/imath] increasing, [imath]x>0[/imath]

[imath]\therefore (1+\tfrac{1}{x})^x[/imath] increasing, [imath]x>0\hspace3ex[/imath] (since [imath]e^x[/imath] increasing)

Now [imath]e^x-x-1[/imath] is strictly increasing, [imath] x>0\hspace2ex (\text{the derivative }e^x-1>0, \;x>0)[/imath]

[imath]\rightarrow e^x-x-1>0, \;x>0 \hspace2ex (\text{since strictly increasing and } e^0-0-1=0)[/imath]

[imath]e^{1/x}-\tfrac{1}{x}-1>0, \;x>0\\ \therefore e^{1/x}>1+\tfrac{1}{x}, \;x>0[/imath]

[imath](1+\tfrac{1}{x})^x<e,\; x>0[/imath]

As for Latex, you're probably best looking at examples that others have done. You should be able to right-click on a formula and show the Latex. However due to a recent upgrade on this site, this doesn't work at the moment. You can use it on other sites though to see their Latex.

You can see the Latex on this site by hitting the reply button however.

Ken