How do you find the limit of #( 1 - (5/x) ) ^x# as x approaches infinity?

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Answer 1 · 2016-06-07T07:35:07

the limit is : #e^-5#

the limit is of the form #1^∞#

# lim_(x->a) (f(x))^(g(x)) = lim_(x->a) ( 1 + ( f(x) - 1 ) )^( g(x) ) #

# lim_(x->a) (f(x))^(g(x)) = e^(lim_(x->a) ( f(x) - 1 ) * (g(x)) #

hence,
In the question given, the limit is of the form : #e^k#

# k = lim_(x->0) (-5/x)*(x) #

# k = -5 #

Therefore, the limit of the above function is : # e^(-5) #