Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{\underset {>}{x \to 0}}\: (\cot \,x)^x=$	A. 1
B. e
C. 0
D.
E.

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{\underset {>}{x \to 0}}\: (\cot \,x)^x=$	A. 1
B. e
C. 0
D.
E. ∞

Denkend aan a^b = e^{b.ln a} kunnen we schrijven :
$\displaystyle\lim_{\underset{>}{x\to 0}}\;(\cot x)^x=\lim_{\underset{>}{x\to 0}}\:e^{x.\ln(\cot x)}\overset{stel}{=}\lim_{\underset{>}{x\to 0}}e^{EXPO}$
We gaan nu die exponent berekenen :
$\\\displaystyle\lim_{\underset{>}{x\to0}}\:{EXPO}=\lim_{\underset{>}{x\to0}}\:x.\ln(\cot x)=\lim_{\underset{>}{x\to0}}\:\frac{\ln(\cot x)}{\frac1x}\left(=\frac{+\infty}{+\infty}\right)\\\buildrel H\over=\lim_{\underset{>}{x\to0}}\:\frac{\frac{1}{\cos x}.\left(-\frac{1}{\sin^2x}\right)}{-\frac{1}{x^2}}=\lim_{\underset{>}{x\to0}}\:\left(\frac{1}{\cot x}.\frac{x}{\sin x}.\frac{x}{\sin x}\right)=\frac{1}{+\infty}.1.1=_{(+)}0$
De gevraagde limiet is dus gelijk aan e⁰ = 1