`lim_(k-gtoo)(k/(k+1))^k`

We can modify the expression inside the bracket to evaluate this. Divide both numerator and denominator by k.

`lim_(k-gtoo)((k/k)/((k+1)/k))^k`

This gives,

`lim_(k-gtoo)(1/(1+1/k))^k`

When we take the limit of this, the expression inside the bracket approaches to `(1/(1+0))` which is 1. Any power of 1, no matter how big is...

## See

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

`lim_(k-gtoo)(k/(k+1))^k`

We can modify the expression inside the bracket to evaluate this. Divide both numerator and denominator by k.

`lim_(k-gtoo)((k/k)/((k+1)/k))^k`

This gives,

`lim_(k-gtoo)(1/(1+1/k))^k`

When we take the limit of this, the expression inside the bracket approaches to `(1/(1+0))` which is 1. Any power of 1, no matter how big is the power it will 1 or close to one. Therefore the given limit approaches to 1.

`lim_(k-gtoo)(1/(1+1/k))^k = 1`

Therefore,

`lim_(k-gtoo)(k/(k+1))^k = 1`