calculus | Gyre&Gimble

Making L’Hôpital’s Rule Go Crazy

Calc II homework problem: Determine ${\lim_{n\rightarrow\infty} \dfrac{n}{\sqrt{n^2+1}}}$ , if it exists.

Answer: This is easy algebra: ${\dfrac{n}{\sqrt{n^2+1}}=\dfrac{\frac{1}{n}n}{\frac{1}{n}\sqrt{n^2+1}}= \dfrac{1}{\sqrt{1+\frac{1}{n^2}}}}$ , which goes to 1 as ${n\rightarrow\infty}$ .

This is nevertheless a Mean Problem. Usually, L’Hôpital’s Rule works on problems like this, but if you try it with this problem, you go into an infinite loop:

${\lim_{n\rightarrow\infty}\dfrac{n}{\sqrt{n^2+1}}= \lim_{n\rightarrow\infty}\dfrac{\sqrt{n^2+1}}{n}= \lim_{n\rightarrow\infty}\dfrac{n}{\sqrt{n^2+1}}= \lim_{n\rightarrow\infty}\dfrac{\sqrt{n^2+1}}{n}=\ldots}$ .

Send to Kindle

math, language and other things that may show up in the wabe

Gyre&Gimble

Tag Archives: calculus

Making L'Hopital's Rule Go Crazy

math, language and other things that may show up in the wabe