/More Calculus And New Glasses

## Are glasses required to be a calculus nerd?

I am learning how to input more and more complex formulas using LaTeX. As I learn more about the formatting options available; fonts, effects, shapes, etc… I can probably use the LaTeX code to design a whole page within a blog post instead of just inserting math equations. There is so much potential for this that I am just beginning to understand.

I’m off to the blood lab today for my preliminary appointment before the big one Friday. Just in the nick of time too because with summer coming to an end I could use some money and I won’t have a lot of free time to donate blood during the fall.

I finally went to the eye doctor yesterday to get a prescription for glasses. This is something I have been meaning to do since January but, better late than never. At least I will be able to see the whiteboards in the classrooms for the fall semester. I still don’t feel comfortable with eyeglasses on but I suppose I’ll get used to them. The free glasses I get from Masshealth are stupid looking so once I figure out how to buy the right ones online I’ll buy ones that I am comfortable wearing, especially since there is Ø chance my eyesight will improve as I get older.

## More LaTeX

L=\lim_{n\to\infty}\Big\{1+\frac{2}{n}\Big\}^n\\lnL=\lim_{n\to\infty}n\cdot\ln\Big\{1+\frac{2}{n}\Big\}\\\textit{using l’hopital’s rule:}\\ln L’=\lim_{n\to\infty} \bigg\{\frac{1}{1+\frac{2}{n}}\bigg\}\cdot\Big\{\frac{-2}{n^2}\Big\}\Big\{\frac{-n^2}{1}\Big\}=2\\e^{lnL’}=e^2\\ L’=e^2\\\textit{by l’hopital’s rule again}\\L=e^2 [\latex]

L=\lim_{n\to\infty}\Bigg \{\sqrt{1+\frac{1}{2n}}\Bigg \}^n = \lim_{n\to\infty}\Bigg \{1+\frac{1}{2n}\Bigg \}^{\frac{n}{2}} \\ \textit{take natural log of both sides}\\lnL=\lim_{n\to\infty}\frac{n}{2} \,ln\Big \{1+\frac{1}{2n}\Big \}=\lim_{n\to\infty}ln \Bigg \{\frac{1+\frac{1}{2n}}{\frac{n}{2}} \Bigg \} \\ \textit{first application of l’hopital’s rule}\\ lnL’=\lim_{n\to\infty}\frac{\Big \{\frac{1}{1+\frac{1}{2n}} \Big\} \cdot \Big\{\frac{-1}{2n^2}\Big \}}{\Big\{\frac{-2}{n^2}\Big \}} \\ \textit{simplify and apply definition of limit}\\ =\lim_{n\to\infty}\frac{1}{4} \cdot\frac {2n}{2n+1}=\frac{1}{4}=lnL’=lnL\\ \mathbf{L=e^{lnL}=e^\frac{1}{4}}

I will be posting my test scores and some sample problems from my Calculus Exams here as soon as the semester is over. It wouldn’t be good to be accused of any impropriety while some students still haven’t taken the second test yet.