Calculate: a) lim x -> infinity (1+2/x)^x You can do this problem in the usualy

Question

Calculate:

a) lim x -> infinity (1+2/x)^x You can do this problem in the usualy way. But you can also do a very clever and quick solution by making the subsitution x=2u

b) lim x -> infinity (cos pi/x)^x

c) lim x -> pi/2- (tanx)^cosx

Explanation / Answer

a.) lim_(x->infinity) (2/x+1)^x Simplify (2/x+1)^x assuming x>0 giving ((x+2)/x)^x: = lim_(x->infinity) ((x+2)/x)^x Indeterminate form of type 1^infinity. Transform using lim_(x->infinity) ((x+2)/x)^x = e^(lim_(x->infinity) x log((x+2)/x)): = e^(lim_(x->infinity) x log((x+2)/x)) Indeterminate form of type 0·infinity. Let t = 1/x, then lim_(x->infinity) x log((x+2)/x) = lim_(t->0) (log((1/t+2) t))/t: = e^(lim_(t->0) (log((2+1/t) t))/t) Indeterminate form of type 0/0. Applying L'Hospital's rule we have, lim_(t->0) (log((2+1/t) t))/t = lim_(t->0) ( d/( dt) log((2+1/t) t))/(( dt)/( dt)): = e^(lim_(t->0) 2/(2 t+1)) Factor out constants: = e^(2 (lim_(t->0) 1/(2 t+1))) The limit of a quotient is the quotient of the limits: = e^(2/(lim_(t->0) (2 t+1))) The limit of 2 t+1 as t approaches 0 is 1: = e^2 b.) lim_(x->infinity) ((cos(pi))/x)^x The limit of ((cos(pi))/x)^x as x approaches infinity is 0: = 0 c.)lim_(x->pi/2-) tan^(cosx) (x) The limit of (tan^(cosx) (x)) as x approaches pi/2- is 1: = 1

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