A confusion in checking differentiability at a point x=1. Cannot find what to write in place of $f(1-h)$ or $f(1+h)$

Question

I am stuck to prove differentiability of a function $f$ at $x=1$.

The given question asks to prove - The limit function of the sequence of functions $\{x^n\}_n$ on $[0,1]$ is differentiable at $x=1$

Here is my trial:-

$$f(x)=\lim_{n \to \infty} f_n(x)={0 \space for \space 0\leq x<1\\=1 \space for \space x=1}$$

Hence $$f(1)=\lim_{n \to \infty} f_n(1)=1$$.

Therefore differentiablility of $f(x)$ at $x=1$,

$$L.H.D.=\lim_{x\to {1-0}} \frac{f(x)-f(1)}{x-1}\\ \Rightarrow\lim_{h\to0} \frac{f(1-h)-f(1)}{(1-h)-1}\\ \Rightarrow\lim_{h\to0} \frac{f(1-h)-1}{h}$$

Is it $0$ or $1$ or $x^n$? - I am not sure which value among these three is to write at $f(1-h)$ or similarly in $f(1+h)$ for $\lim_{h\to0}$ to determine the lhd and rhd.

Here, I need help and any help is highly appreciated.

Answer 1

The $f$ is not even continuous at $x=1$, so it is not differentiable. Note that $\lim_{h\rightarrow 0}\dfrac{f(x)-f(1)}{x-1}=\lim_{h\rightarrow 0}\dfrac{f(1-h)-f(1)}{-h}=\lim_{h\rightarrow 0}\dfrac{f(1+h)-f(1)}{h}$.

Answer 2

For $$x\in [0,1), f (x)=\lim_{n\to+\infty}x^n=0$$

for $x=1, f (1)=\lim_{n\to+\infty}f_n(1)=1.$

the limit function is not continuous at $x=1$ and therefore not differentiable.

The convergence is not uniforme at $[0,1] $.