The value of k for which $f(x)=\left\{\begin{array}{cc} \frac{x^{2^{32}}-2^{32} x+4^{16}-1}{(x-1)^2} & , x \neq 0 \\ k & , x=1 \end{array}\right.$

is continuous at x = 1, is

$2^{63}-2^{31}$

If f(x) is continuous at x = 1, then

$\lim\limits_{x \rightarrow 1} f(x)=f(1)$

$\Rightarrow \lim\limits_{x \rightarrow 1} f(x)=k$

$\Rightarrow \lim\limits_{x \rightarrow 1} \frac{x^{2^{32}}-2^{32} x+4^{16}-1}{(x-1)^2}=k$

$\Rightarrow \lim\limits_{x \rightarrow 1} \frac{\left(x^n-n x+n-1\right)}{(x-1)^2}=k$,  where  $n=2^{32}$

$\Rightarrow \lim\limits_{x \rightarrow 1} \frac{\left(x^n-1\right)-n(x-1)}{(x-1)^2}=k$

$\Rightarrow \lim\limits_{x \rightarrow 1} \frac{\frac{x^n-1}{x-1}-n}{x-1}=k$

$\Rightarrow \lim\limits_{x \rightarrow 1} \frac{\left(x^{n-1}+x^{n-2}+...+x+1\right)-n}{x-1}=k$

$\Rightarrow \lim\limits_{x \rightarrow 1} \frac{\left(x^{n-1}-1\right)+\left(x^{n-2}-1\right)+...+(x-1)}{x-1}=k$

$\Rightarrow (n-1)+(n-2)+...+2+1=k$

$\Rightarrow k=\frac{n(n-1)}{2}=\frac{2^{32}\left(2^{32}-1\right)}{2}=2^{63}-2^{31}$