Wiskunde Multiple Choice Vraag

$\displaystyle\lim_{x\to 0}\left(\frac{\sin x}{x}\right)^{\frac{1}{1-\cos x}}=$	A. \(e\)
B. \(e^{-1}\)
C. \(-\sqrt[3]e\)
D. \(\frac{1}{\sqrt[3]e}\)
E. \(-\sqrt e\)

$\inline\displaystyle\lim_{x\to 0}\left(\frac{\sin x}{x}\right)^{\frac{1}{1-\cos x}}=\lim_{x\to 0}\left(1+\frac{\sin x}{x}-1\right)^{\frac{1}{1-\cos x}}\\=\left[\lim_{x\to 0}\left(1+\frac{\sin x-x}{x}\right)^{\frac{x}{\sin x-x}}\right]^\frac{\sin x-x}{x(1-\cos x)}\\=\textbf{e}\text{tot de macht}\lim_{x\to 0}\frac{\sin x-x}{x-x\cos x}=\textbf{L}$

$\inline\\\textbf{L}\left(=\frac00\right)=\displaystyle\lim_{x\to 0}\frac{\cos x-1}{1-\cos x+x\sin x}\left(=\frac00\right)=\lim_{x\to 0}\frac{-\sin x}{\sin x+\sin x+x\cos x}\\\left(=\frac00\right)=\lim_{x\to 0}\frac{-\cos x}{\cos x+\cos x+\cos x-x\sin x}=\frac{1}{1+1+1-0}=-\frac13$ Het antwoord is bijgevolg $e^{-\frac13}=\frac{1}{\sqrt[3]e}$

^{gricha - vA110 - 1.1.2026}