Limits 1

I am not sure, so I expect to be corrected. First, $\lim _{ n\rightarrow \infty }{ \sqrt [ n ]{ p } -\sqrt [ n ]{ q } } =1-1=0$ . So $\sqrt [ n ]{ p }$ tends to be equal to $\sqrt [ n ]{ p }$ . Then, applying MA-MG: $\lim _{ n\rightarrow \infty }{ { \left( \frac { \left( \sqrt [ n ]{ p } +\sqrt [ n ]{ q } \right) }{ 2 } \right) }^{ n } } ={ \lim _{ n\rightarrow \infty }{ { \left( \sqrt { \sqrt [ n ]{ pq } } \right) }^{ n } } =\lim _{ n\rightarrow \infty }{ \sqrt { pq } } }=\sqrt { pq }$

Note: I am pretty sure that I made a mistake

lim n to infinity ([ $\frac{[p^(1/n) + q^(1/n)]}{2}$ )^n

1+ ( $\frac{p^(1/n) + q^(1/n)-2}{2}$ )^n using lim x to infinity (1+1/x)^x tends to e. e^ lim n to infinity( $\frac{(p^(1/n)-1 + q^(1/n)-1)n}{2}$ )

e^ lim n to infinity $\frac{p^(1/n).logp + q^(1/n)log q}{2}$

Finally, we get (pq)^1/2

This can be easily solved using the options. First let's put p=q=1.From this we get 1this eliminates option 4.next let us put p=q. Doing so we get p as the answer. The only option which gives p as the answer when you put p=q is option 1