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Find the limit of the sequence

$ \left\{ \sqrt 2, \sqrt{2\sqrt2}, \sqrt{2\sqrt{2\sqrt2}}, \cdot \cdot \cdot \right\} $

the sequence increases, so $L=2$

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Let's find the limit of this sequence. If we look at the first term, weaken right, This is too too. The one half the second star weaken right. This is two times two to the one half to the one half power. So simplifying this we have to do the three forints. And for the third term, let's do this as well. This is two times two to the three fourths to the one half. So that's two to the seven over eight. And it looks like we may have enough here to see the pattern. So the end of term is given by a N equals two. And now what's the exponents? So we had one half three fourth, seven eight. So notice that the numerator is always one less than the denominator and noticed that the denominator is of the form to the end. So this means we have to to the end and then numerator is one less. So you go back up there, subtract one, and so we have limit, take a luminous and goes to infinity. So now we have to to the end power minus one groups a minus one should not be in the exponents. It should be to the whole thing and then to the end. Now we'LL use the fact that this power function here, this exponential function excuse me is continuous so I can go in and take this limit and right in the exponents. Yeah, and then either by doing some algebra or using low Patel's rule, you can see that this exponents goes toe one. So this is just to the one which equals two. And this is the limit of the sequence to, and that's our final answer.