Limit for $(1-\cos x)^{(k+x)}/x$ when $x \to 0$ Without the exponent, the fraction would be 0 as $\lim x \to 0$. How do you handle it when there is an exponent?

Since $(1-\cos x) ^{x} \to 1$ as $x\to 0$ (prove this!) the desired limit is equal to the limit of $(1-\cos x)^{k} /x$ as $x\to 0$. Further note that $(1-\cos x) /x^{2}\to 1/2$ hence the desired limit is equal to the limit of $2^{-k}x^{2k-1}$. Thus the desired limit is equal to $1/\sqrt{2}$ if $k=1/2$ and it is $0$ if $k>1/2$ and diverges if $k<1/2$. All this is valid for $x\to 0^{+}$. When we take into account $x\to 0^{-} $ then the limit is $-1/\sqrt{2}$ for $k=1/2$.
Thus to conclude, the limit does not exist if $k\leq 1/2$ and is $0$ if $k>1/2$.
