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Making L'Hopital's Rule Go Crazy

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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}.

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3 thoughts on “Making L'Hopital's Rule Go Crazy”

  2. Incidentally, if we can in some other way establish that the limit exists — e.g. the approximand is increasing and bounded above — we can see from l’Hopital that it is its own reciprocal, and thus must be 1. This reminds me of integrating by parts with trigonometric functions.

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