Convergence of sequence of functions $ f_n(x)=\begin{cases} n^{p+1}x & \text{if $0≤x≤1/n$}\\ \frac{1}{x^p} & \text{if $1/nQuestion: Let $p$ be positive real number and consider the sequence $\{f_n\}_{n=1}^\infty$ defined by
$$
f_n(x)=\begin{cases}
         n^{p+1}x & \text{if $0≤x≤\displaystyle\frac1n$}\\
        \displaystyle\frac{1}{x^p} & \text{if $\displaystyle\frac1n<x≤1$}
     \end{cases}
$$
Let $f(x)=\lim_{n\to\infty} f_n(x)$ where $x\in[0,1].$
Then
(i) find $f(x)$
(ii) Is $f(x)$ is Riemann integrable?
(iii) Is the convergence uniform?
My attempt: As $n\to\infty$ we have $1/n\to0$ and hence $x=0$. So I thought $f(x)=\lim_{n\to\infty} f_n(x)=0$ and hence $f$ is Riemann integrable. But, key says I am wrong. How to find $f(x)$? Please help.

Notice that for $n$ large enough, $1/n\to 0$ so your function $f(x)=\frac{1}{x^p}$ for $0< x\leq 1$. In $0$ the function should be define as zero. So we have a function which is not bounded in its interval of definition so couldn't be Riemann integrable, for $p\geq 1$. For the case $0<p<1$ we have that the improper $\int_0^1 \frac{1}{x^p}dx$ converges (not really Riemann integrable, but I think for your purposes could be say yes it is.) Notice that for uniform convergence we should normally have $$ \lim_{n\to\infty} \sup_{x\in[0,1]}|f(x)-f_n(x)|=0$$ but in this case, each $f_n$ is bounded, but $f(x)$ is not bounded, it means that $$\lim_{n\to\infty} \sup_{x\in[0,1]}|f(x)-f_n(x)|=\infty$$ and the convergence is not uniform. Indeed, faster: as each $f_n$ is Riemann integrable, but its limit $f$ is not, then the convergence is not uniform.
