#y_n=log x_n, n =2,3,4,...and y_n(n1)/n y_(n1)=1/n log n#, with #y_2=log sqrt2#, how do you prove that #x_n=(n!)^(1/n)#?
By induction
Proof: (By induction)
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To prove that ( x_n = (n!)^{1/n} ) given the recursive relation ( y_n = \log{x_n} ) for ( n = 2, 3, 4, \ldots ) and ( y_n  \frac{n1}{n} y_{n1} = \frac{1}{n} \log{n} ), with ( y_2 = \log{\sqrt{2}} ), we can proceed as follows:

We start by rewriting the given recursive relation: [ y_n  \frac{n1}{n} y_{n1} = \frac{1}{n} \log{n} ]

Since ( y_n = \log{x_n} ), we substitute this into the equation: [ \log{x_n}  \frac{n1}{n} \log{x_{n1}} = \frac{1}{n} \log{n} ]

Taking exponentials of both sides, we have: [ x_n = x_{n1}^{\frac{n1}{n}} \cdot n^{\frac{1}{n}} ]

Now, we apply the initial condition ( y_2 = \log{\sqrt{2}} ), which gives us ( x_2 = \sqrt{2} ).

We can now use the recursive relation to find subsequent values of ( x_n ): [ x_3 = x_2^{\frac{2}{3}} \cdot 3^{\frac{1}{3}} ] [ x_4 = x_3^{\frac{3}{4}} \cdot 4^{\frac{1}{4}} ] [ \vdots ] [ x_n = x_{n1}^{\frac{n1}{n}} \cdot n^{\frac{1}{n}} ]

By induction, we can show that for any ( n ), ( x_n = (n!)^{1/n} ) holds.
Thus, by recursively applying the given relation and using the initial condition, we can prove that ( x_n = (n!)^{1/n} ).
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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