The function   $f\left(x\right)=\left\{\begin{array}{l}1-2x+3x^2-4x^3+\mathrm{........},\left|x\right|<1\\ 1,x=-1\end{array}\right.is$ 

Since   $f\left(x\right)=1-2x+3x^2-4x^3+\mathrm{.........}\infty =\frac{1}{{\left(1+x\right)}^2}$

$Atx=-1$

$\underset{x\to -1}{\lim }f\left(x\right)=\underset{x\to -1}{\lim }\frac{1}{{\left(1+x\right)}^2}=\underset{x\to 0}{\lim }\frac{1}{{\left(1-1+h\right)}^2}=\infty$

But   $f\left(-1\right)=1$

$\therefore f\left(x\right)isnotcontributesatx=-1$

$Thisimpliesf\left(x\right)isnotderivableatx=-1$