I have this function, and I want to take the limit of it. On the open interval (0,1), the limit is 1, but on the closed interval, when you fix $x$ to be equal to 1, then the limit turns into $\frac{1}{2}$. Does that mean there is no limit on the closed interval?
Also, I'm trying to show that the function converges uniformly for $0 < a < 1$, on the closed interval [0, a]. I'm having trouble with how $a$ is not on the interval, but the closed interval is unif. convergent. I understand the need for the "sup-criterion" but I'm struggling to find the supremum that goes to 0 as $n \to \infty$.
Then I need to show that $f_n$ does not converge uniformly to $f$ on [0,1]. My guess is that using the supremum above, the inclusion of 1 makes the limit non-zero and thus the function won't converge uniformly. Might be able to use counter-example too, rather unsure.
For any $\;a\in [0,1)\;$ , we have that
$$\frac1{1+a^n}\xrightarrow[n\to\infty]{}\frac1{1+0}=1$$
For $\;a=1\;$ the limit is, as you correctly stated, $\;\frac12\;$
Thus, the convergence cannot be uniform on $\;[0,1]\;$ since, as all the functions $\;f_n\;$ are continuous, then also the uniform limit function would be continuous, and it isn't....