As for solving in a few seconds, it certainly took me less than 13 seconds: for anyone familiar with Taylor expansions, and using that 1-cos^x=sin^2x, the limit is the same as the limit

(-x-x)/x^2=-2/x,

and the limit at zero doesn’t exist. Similarly, the “mistaken” limit with a product instead of a minus would be

-xx/x^2=-1,

even faster than the original version.

]]>lim [1/(1-x) – cos x] / sin(2x) , which is still indeterminate.

Another iteration (again forgetting chain) leaves lim [-1/(1-x)^2 + sin x] / cos(2x) = -1

This computation can be done within the time constraints given in the film.

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