`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 Now
Start 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`