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For any $n \in \Bbb N$, let $f_n:\Bbb R \to \Bbb R$ be a function defined by $$f_n(x)=\frac{nx^3}{7+nx^2}, \quad \forall x \in \Bbb R.$$ Investigate whether the sequence $(f_n)$ is uniformly convergent on $\Bbb R$.

attempt: I've shown that $f_n \to f$ on $\Bbb R$, where $f:\Bbb R \to \Bbb R$ is a function defined by \begin{equation*} f(x)= \begin{cases} 0, \quad x=0 \\ x, \quad x\ne 0. \end{cases} \end{equation*}

Notice that for any $n \in \Bbb N$ and for any $x \in \Bbb R \setminus \{0\}$, we have $$\frac{x^2}{(7+nx^2)^2} < \frac{x^2}{7+nx^2} < \frac{x^2}{nx^2} \implies \frac{|x|}{7+nx^2}<\frac{1}{\sqrt{n}}.$$ I claimed that $(f_n)$ is converges uniformly on $\Bbb R$ to $f$. To this end, let $\epsilon>0$ be arbitrary. Choose $K \in \Bbb N$ with $K>\frac{49}{\epsilon^2}$ such that for any $n \in \Bbb N$ with $n \ge K$, we have $$|f_n(x)-f(x)| = \left|\frac{nx^3}{7+nx^2}-x \right| = \frac{7|x|}{7+nx^2} < \frac{7}{\sqrt{n}} \le \frac{7}{\sqrt{K}}< \epsilon,$$ for any $x \in \Bbb R\setminus \{0\}$. Hence, $(f_n)$ is converges uniformly on $\Bbb R$ to $f. \qquad \Box$

Does it correct? On the other hand, I think if the sequence of functions converges to a piecewise function, then the sequence is not uniformly convergent. I confusing on it. What's is the correct approach? Thanks in advanced.

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  • $\begingroup$ note that $f$ is continuous $\endgroup$
    – ling
    May 14 at 1:26
  • $\begingroup$ @ling So? Does it correct or not? $\endgroup$
    – gerrr
    May 14 at 1:28
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    $\begingroup$ I think it is correct. No need for f to be piece-wise defined. f(x) = x. $\endgroup$
    – 311411
    May 14 at 1:31
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    $\begingroup$ Your approach is correct . $\endgroup$ May 14 at 1:59

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