Mathematical insult gone awry

Seen on average image board:

math_fail

Let’s try to solve it. The inner limit can be written as

\lim\limits_{x\rightarrow 0} e^{\frac{\ln(1+\sin^2x + \sin^2 2x + \dots+ \sin^2nx)}{x^2}}

The limit in the exponent can be taken using the l’Hostpital rule

\lim\limits_{x\rightarrow 0} \frac{\ln(1+\sin^2x + \sin^2 2x + \dots+ \sin^2nx)}{x^2} = \lim\limits_{x\rightarrow 0} \frac{\sin x \cos x+2\sin 2x \cos 2x + \dots + n\sin nx \cos nx}{x(1+\sin^2x + \sin^2 2x + \dots+ \sin^2nx)} = \dots

The in the sense of the limit the denominator can be replaced by x. In the same sense we can put any \cos \alpha x \rightarrow 1 in the numerator. Any expression of the form \sin\alpha x/x \rightarrow \alpha as x\rightarrow 0, thus

\dots = 1 +2\cdot 2 + \dots + n\cdot n = \frac{n^3}{3} + \mathcal{O}(n^2)

The rest is straightforward and one immediately sees that the limit is fact

e^{\frac{1}{3}}

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About goobypl5

pizza baker, autodidact, particle physicist
This entry was posted in Math and tagged , . Bookmark the permalink.

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