copy and paste this google map to your website or blog!
Press copy button and paste into your blog or website.
(Please switch to 'HTML' mode when posting into your blog. Examples: WordPress Example, Blogger Example)
Input Form:Deal with this potential dealer,buyer,seller,supplier,manufacturer,exporter,importer
(Any information to deal,buy, sell, quote for products or service)
Prove $\ln (x)$ is continuous using $\epsilon-\delta$ Remark: Actually, $1-\frac {1} {e^ {\epsilon}}$ is the smaller of the two, so in effect we are letting that be $\delta$ But we really don't need to bother finding that out: all we need to do is to show there is a $\delta$ that works
real analysis - Does the epsilon-delta definition of limits truly . . . This quote resonates with my current dilemma Does the Epsilon-Delta definition truly capture the essence of what we mean by a 'limit'? though the epsilon-delta definition is a mathematical construct, what evidence do we have that it accurately reflects our intuitive concept of a limit?
Clarification on bounding $\frac {1} {|x-3|}$ in $\epsilon$-$\delta . . . To guarantee that it is less than $\epsilon$, you need to have some condition on $|x-5|$ itself in terms of $\epsilon$ This is the thing you wrote in the final line The full argument written properly must start from $|x-5|<\min (1,2\epsilon)$ to somehow conclude $\left|\frac {1} {x-3}-\frac {1} {2}\right|=\frac {|x-5|} {2|x-3|}<\epsilon$