If `a_(n)` and `b_(n)` are positive integers and `a_(n)+sqrt2b_(n)=(2+sqrt2))^(n)`, then `lim_(nrarroo) ((a_(n))/(b_(n)))`=A. `sqrt2`B. 2C. `e^(sqrt2)`D. `e^(2)`

If `a_(n)` and `b_(n)` are positive integers and `a_(n)+sqrt2b_(n)=(2+

1.	If `a_(n)` and `b_(n)` are positive integers and `a_(n)+sqrt2b_(n)=(2+sqrt2))^(n)`, then `lim_(nrarroo) ((a_(n))/(b_(n)))`=A. `sqrt2`B. 2C. `e^(sqrt2)`D. `e^(2)`
Answer» Correct Answer - A We have `a_(n)+sqrt2b_(n)=(2+sqrt2)^(n)` `rArr" "a_(n)-sqrt2b_(n)=(2-sqrt2)^(n)` Solving we get `a_(n)=(1)/(2)[(2+sqrt2)^(n)+(2-sqrt2)^(n)]` and `b_(n)=([(2+sqrt2)^(n)+(2-sqrt2)^(n)])/([(2+sqrt2)^(n)-(2-sqrt2)^(n)])` `" "=sqrt2([1+((2-sqrt2)/(2+sqrt2))^(n)])/([1-((2-sqrt2)/(2+sqrt2))^(n)])` Hence, `underset(nrarroo)(lim)((a_(n))/(b_(n)))=sqrt2((1+0)/(1-0))=sqrt2" "(because(2-sqrt2)/(2+sqrt2)lt1)`

`lim_(xto oo) (1)/(n) {1+e^(1//n)+e^(2//n)+e^(3//n)+.....+e^((n-1)/(n))}` is equal toA. eB. `-e`C. `e-1`D. `1-e`
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