Find the value of this limit. Give the solution in details please.
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Replacing n with
would result in an indeterminant form. So we use l hopital's rule where we take derivative of both numerator and denominator.
Now the new limit would ask for 2^(n)/2^(n) as n approaches infinity. The expression is 1. so the limit is simply 1.
would result in an indeterminant form. So we use l hopital's rule where we take derivative of both numerator and denominator.
Now the new limit would ask for 2^(n)/2^(n) as n approaches infinity. The expression is 1. so the limit is simply 1.
mr1ceres:
Would you please answer this question by not using l hospital rule. and replacing n with other varialbe and making limit tending to 0.
Ok....here's another way for it. lim(n->infinity) (2^n -1)/(2^n +1) . Substitute 2^n=y. As n approaches infinity, y also approaches infinity. Now, lim(y->infinity) (y-1)/(y+1). lim(y->infinity) (y(1-1/y))/(y(1+1/y)). lim(y->infinity) (1-1/y)/(1+1/y). As y approaches infinity, 1/y approaches 0. Therefore the limit equals (1-0)/(1+0)=1.
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