CAPTCHAs are designed to make you prove you are a human and to stop a computer program from contributing to comment spam or other bad behaviors. Neatorama has a link to some extremely difficult captchas that make you prove you are super human.

One of my favorites is:

Firstly, I assume they want the solution to the equation in the box, my Russian is non-existent. The equation isn't too difficult once you parse it. I think that the limit is ln(2), because arctan(x) goes to 0 while sin(1/x) can only bounce back and forth wildly between -1 and 1 as x grows. You can see that in the graph below (click for larger).

However, the square root won't like the parts of the function when sin(1/x) is negative, but if complex numbers using

*i*are allowed and we make this a complex function you would still get to ln(2) eventually. The graph shows these blank areas on every oscillation as x gets smaller and smaller as you move left on the plot.Probably we should expand the functions into Taylor series and then do the limit (maybe using L'Hopital's rule also) to get a more rigorous result. Coincidentally, it also goes to ln(2) when x goes to infinity, this time because sin(1/x) goes to zero, while the arctan of x is bounded between -pi and +pi.

## 2 comments:

Решение:

lim (x->0) ln (1 - cos /ln tg x = lim (x->0) (ln (1 - cos )'/(ln tg ' =

= lim (x->0) (sin x/(1 - cos )/(1/cos^2 x * 1/tg =

= lim (x->0) (sin x/(1 - cos )/(1/(sin x * cos ) =

= lim (x->0) (sin^2 x * cos /(1 - cos = cos 0 * lim (x->0) sin^2 x/(1 - cos =

= lim (x->0) sin^2 x/(1 - (1 - 2 * sin^2 (x/2))) =

= lim (x->0) sin^2 x/(2 * sin^2 (x/2)) = lim (x->0) x^2/(2 * (x/2)^2) = lim (x->0) x^2/(x^2/2) = 2.

Но в примере ошибка в sin 1/x так как x->0

Решение:

lim (x->0) ln (1 - cos /ln tg x = lim (x->0) (ln (1 - cos )'/(ln tg ' =

= lim (x->0) (sin x/(1 - cos )/(1/cos^2 x * 1/tg =

= lim (x->0) (sin x/(1 - cos )/(1/(sin x * cos ) =

= lim (x->0) (sin^2 x * cos /(1 - cos = cos 0 * lim (x->0) sin^2 x/(1 - cos =

= lim (x->0) sin^2 x/(1 - (1 - 2 * sin^2 (x/2))) =

= lim (x->0) sin^2 x/(2 * sin^2 (x/2)) = lim (x->0) x^2/(2 * (x/2)^2) = lim (x->0) x^2/(x^2/2) = 2.

Но в примере ошибка в sin 1/x так как x->0

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