`a_n = (2n)/sqrt(n^2+1)` Find the limit (if possible) of the sequence.

`a_n=(2n)/sqrt(n^2+1)`

To find the limit of a sequence, let n approach infinity.

`lim_(n->oo) a_n`

`=lim_(n->oo) (2n)/sqrt(n^2+1)`

`= lim_(n->oo) (2n)/sqrt(n^2(1+1/n^2))`

`=lim_(n->oo) (2n)/(nsqrt(1+1/n^2))`

`=lim_(n->oo) 2/sqrt(1+1/n^2)`

`=(lim_(n->oo)2)/(lim_(n->oo)sqrt(1+1/n^2))`

`=2/sqrt(1+0)`

`=2`

Therefore, the limit of the sequence is 2.

Comments

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