Answer to Question #91436 in Real Analysis for Kiran

Question #91436
Show that limit n approaches to infinity (xe^-nx)=0 for x € real number , x>0
1
Expert's answer
2019-07-05T12:38:15-0400

Let "a = e^{-x}", "x>0," then "a \\in (0,1)". Then It is well-known that


"\\lim_{n\\to \\infty} a^n = 0."


This fact can be proved by the definition of the limit: in order that "a^n < \\varepsilon" it is sufficient that "n > \\log_a \\varepsilon".

Thus,


"\\lim_{n\\to \\infty} xe^{-nx} = x\\lim_{n\\to \\infty} (e^{-x})^n = x \\lim_{n\\to \\infty} a^n = x \\cdot 0= 0."

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