# Find Limit: Where Limit ('k' approaches Infinity) [ k/(k+1) ]^k ?Problem is how to simplify the algebraic expression so that its limit can be calculated !!

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### 1 Answer

`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`