26: Since the sine in the denominator is at most one in absolute value, it is negligible compared to n² when n gets large and can be ignored. Similarly in the numerator, cosine is at most one, so the term ncos(n) is (in absolute value) at most n and thus can be ignored compared to 2n², clearly one can also ignore the "−1". So one gets

How to prove this guess that the given sequence goes to 2? Since we guessed that both the numerator and the denominator are in fact of type n², they should tend to infinity. Thus one could try to use the l'Hospital rule. However, this would not be the best idea. First, one would have to prove that the numerator and denominator go to infinity in order to justify its use, and that would mean extra work. Second, derivative does not get rid of sine and/or cosine, so things would not become easy; perhaps more l'Hospital's would help, perhaps not even that.

Another option is to use the knowledge of dominant powers, cancel n² in the fraction and show that the resulting non-constant parts go to zero. In particular, one would have to show that the cosine divided by n and the sine divided by n² go to zero. This is actually easy using the Squeeze theorem, in fact this is a typical school problem. So this way it should work.

In this procedure, one would have to use the squeeze twice. With a little bit of luck, just one squeeze will be enough when applied to the whole given fraction. Try it.

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