 # Limit for $$\displaystyle\frac{{\left({1}-{\cos{{x}}}\right)}^{{{\left({k}+{x}\right)}}}}{{x}}$$ when $$\displaystyle{x}\to{0}$$ aanvarendbq28 2022-03-23 Answered
Limit for $\frac{{\left(1-\mathrm{cos}x\right)}^{\left(k+x\right)}}{x}$ when $x\to 0$
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Hint:
$\frac{{\left(1-\mathrm{cos}x\right)}^{\left(\frac{32}{\pi }+x\right)}}{x}={\left(1-\mathrm{cos}x\right)}^{\left(\frac{32}{\pi }+x-1\right)}\left(\frac{1-\mathrm{cos}x}{x}\right)$
###### Not exactly what you’re looking for? Nyla Trujillo
Since ${\left(1-\mathrm{cos}x\right)}^{x}\to 1$ as $x\to 0$ (prove this!) the desired limit is equal to the limit of $\frac{{\left(1-\mathrm{cos}x\right)}^{k}}{x}$ as $x\to 0$. Further note that $\frac{1-\mathrm{cos}x}{{x}^{2}}\to \frac{1}{2}$ hence the desired limit is equal to the limit of ${2}^{-k}{x}^{2k-1}$. Thus the desired limit is equal to $\frac{1}{\sqrt{2}}$ 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 $-\frac{1}{\sqrt{2}}$ for k=1/2.
Thus to conclude, the limit does not exist if $k\le \frac{1}{2}$ and is 0 if k>1/2