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Let us consider the sequence of function $$ f_n(x)= \begin{cases} nx, x\in [0, 1/2^n] \\ 1/nx, x \in (1/2^n, 1]\\ \end{cases} $$ Find the limit function $f(x).$

My work:

if $x \in [0, 1/2^n]$ then $f_n(x)$ goes to zero for sufficiently large $n.$ For $x = 1,$ $f_n(x)$ tends to $0$ for $n \rightarrow \infty.$ Now for $1/2^n<x<1,$ keep $x$ fixed. Then $1/2^n < x$ for sufficiently large $n.$ Now what can I conclude about the limit function $f(x)?$

Shootforthemoon
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Manglu
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1 Answers1

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You know that $1/2^n \underset{n \rightarrow +\infty}{\rightarrow}0$ so for all $\epsilon>0$, it exists a rank $N$ so that $1/2^n<\epsilon$ then $f_n$ tends to be $x \mapsto 1/nx$ which converges to $0$.

Atmos
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