Which sequence of functions converges uniformly I want to check which of the following sequences of functions converges uniformly.

*

*$\sin (x^n), \ x\in\left[\frac{1}{2}, 1\right ]$

*$\sin (x^{\frac{1}{n}}), \ x\in\left[0, \frac{1}{2}\right ]$

*$\sin (x^{\frac{1}{n}}), \ x\in\left[\frac{1}{2}, 1\right ]$

*$\sin (x^n), \ x\in\left[0,\frac{1}{2}\right ]$
I have done the following :
First we check the pointwise convergence, so we consider the limit as $n\rightarrow \infty$.
We have the following:
$$\lim_{n\rightarrow \infty}\sin (x^n)=0 \text{ for both intervals of } x $$
$$\lim_{n\rightarrow \infty}\sin (x^{\frac{1}{n}})= \sin (1) \text{ for both intervals of } x$$
Then we have to consider the functions $g_1(x)=|\sin (x^n)-0|=|\sin (x^n)|$ and $g_2(x)=|\sin (x^{\frac{1}{n}})-\sin (1)|$ and to check if the minimum goes to $0$ as $n\rightarrow \infty$, right?
For that need the derivatives of these two functions, or not?
We have $g_1'(x)=|nx^{n-1}\cos (x^n)|$ and $g_2'(x)=|\frac{1}{n}x^{\frac{1}{n}-1}\cos (x^{\frac{1}{n}}))|$.
Then we need the roots of the first derivative, right?
Or do we have to do that in an other way?
 A: Let $f_n(x)=\sin(x^n)$ and $g_n(x)=\sin(x^{\frac{1}{n}})$.
We will also use:
$$f(x)=\left\{ \begin{array}{ll}
      0 \text{ if } x\in [0,1[ \\
      \sin (1) \text{ if } x=1 \\
\end{array} \right.$$
$$g(x)=\left\{ \begin{array}{ll}
      0 \text{ if } x=0 \\
      \sin (1) \text{ if } x \in ]0,1] \\
\end{array} \right.$$
Pointwise convergence
Let $x \in [0,1[$. Then, $\underset{n \rightarrow +\infty}{\lim} f_n(x)=\sin(0)=0$, because $\sin$ is continuous. For $x=1$, we have $\underset{n \rightarrow +\infty}{\lim} f_n(1)=\underset{n \rightarrow +\infty}{\lim} \sin(1)=\sin(1)$. Therefore, $(f_n)_{n \in \mathbb{N}}$ converges to $f$ on $[0,1]$.
Let $x \in ]0,1]$. Then, $\underset{n \rightarrow +\infty}{\lim} g_n(x)=\sin(1)$. For $x=0$, $\underset{n \rightarrow +\infty}{\lim} g_n(0)=\underset{n \rightarrow +\infty}{\lim} 0=0$. So, $(g_n)_{n \in \mathbb{N}^*}$ converges to $g$ on $[0,1]$.
Uniform convergence

*

*For all $n \in \mathbb{N}$, $f_n$ is continuous on $[\frac{1}{2},1].$ Since $f$ isn't continuous at $1$, we can conclude that $(f_n)_{n \in \mathbb{N}}$ doesn't converge uniformly to $f$ on $[\frac{1}{2},1]$.

*For all $n \in \mathbb{N}^*$, $g_n$ is continuous on $[0,\frac{1}{2}].$ Since $g$ isn't continuous at $0$, we can conclude that $(g_n)_{n \in \mathbb{N}^*}$ doesn't converge uniformly to $g$ on $[0,\frac{1}{2}]$.

*$\forall x \in [\frac{1}{2},1], |g_n(x)-g(x)|=|\sin(x^{\frac{1}{n}})-\sin(1)| \leq |\sin(\frac{1}{2}^{\frac{1}{n}})-\sin(1)| \underset{n \rightarrow +\infty} {\rightarrow} 0$. Consequently, $\underset{n \rightarrow +\infty}{\lim} \underset{x \in [\frac{1}{2},1]}{\sup} |g_n(x)-g(x)|=0$ and $(g_n)_{n \in \mathbb{N}^*}$ converges uniformly to $g$ on $[\frac{1}{2},1]$.

*$\forall x \in [0,\frac{1}{2}], |f_n(x)-f(x)|=|\sin(x^n)-0| \leq |\sin(\frac{1}{2}^n)| \underset{n \rightarrow +\infty} {\rightarrow} 0$. Consequently, $\underset{n \rightarrow +\infty}{\lim} \underset{x \in [0,\frac{1}{2}]}{\sup} |f_n(x)-f(x)|=0$ and $(f_n)_{n \in \mathbb{N}}$ converges uniformly to $f$ on $[0,\frac{1}{2}]$.

Hope it helps.
