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?
 A: Hint: $$\frac{(1-\cos x)^{(\frac32 \pi+x)}}{x} = (1-\cos x)^{(\frac32 \pi+x-1)}\left(\frac{1-\cos x}{x}\right)$$
A: 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 $k>1/2$.
A: Considering $$A=\frac{(1-\cos x)^{(k+x)}}x\implies \log(A)=(k+x) \log (1-\cos (x))-\log (x)$$ use Taylor series $$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^6\right)$$ $$1-\cos(x)=\frac{x^2}{2}-\frac{x^4}{24}+O\left(x^6\right)$$ $$\log (1-\cos (x))=(2 \log (x)-\log (2))-\frac{x^2}{12}+O\left(x^4\right)$$
$$(k+x) \log (1-\cos (x))=k (2 \log (x)-\log (2))+x (2 \log (x)-\log (2))+O\left(x^2\right)$$ $$\log(A)=((2 k-1) \log (x)-k \log (2))+x (2 \log (x)-\log (2))+O\left(x^2\right)$$ making, close to $x=0$,
$$A \approx \frac{x^{2k-1}}{2^k}$$ from which you can conclude.
