I was wondering how to prove or disprove the following sequence of functions is uniformly convergent
$$\ f(n) = \frac{nt}{nt+1}, n≥1, t:[0,1] \to R$$
So far I have analyzed the limits at $t=0$ and $t=1$ and believe it to be point-wise convergent, but not uniformly convergent. However, I'm not sure how to prove this.
Also, how does the above sequence vary from
$$\ f(n) = \frac{nt}{n+t}, n≥1, t:[0,1] \to R$$ in regards to uniform convergence?
On
For $t$ fixed in $(0,1]$,
$$\lim_{n\to+\infty}f_n(t)=1=f(t)$$
$$\lim_{n\to+\infty}f_n(0)=0=f(0)$$
the pointwise limit $f$ is not continuous at $[0,1]$, the convergence is Not uniform since all $f_n$ are continuous at $[0,1]$.
On
The sequence
$$f_n(t) = \frac{nt}{nt+1}$$
converges (in a point-wise sense) to:
$$f(t) = \begin{cases} 0 & \text{if}~t=0\\ 1 & \text{if}~0<t\leq 1 \end{cases}.$$
To check uniform convergence, you need to evaluate $d_n(t) = |f_n(t) - f(t)|$:
$$d_n(t) = \begin{cases} 0 & \text{if}~t=0\\ \displaystyle \frac{1}{nt+1} & \text{if}~0<t\leq 1 \end{cases}.$$
Notice that:
$$\sup_{t \in [0,1]} d_n(t) = 1,$$
which does not converge to $0$ as $n$ goes to infinity. Therefore, convergence is not uniform.
Observe $$f_n(t)=\frac{nt}{nt+1}=\dfrac{t}{t+\dfrac{1}{n}}\to\begin{cases}0 , &t=0\\1, &t\neq0\end{cases}$$