\require{AMSmath} Voorbeeld 1 Gevraagd: $\eqalign{\mathop {\lim }\limits_{x \to 0} x^2 \sin \left( {\frac{1}{x}} \right)}$ $\eqalign{ & \mathop {\lim }\limits_{x \to 0} x^2 \sin \left( {\frac{1}{x}} \right) \cr & - x^2 \leq x^2 \sin \left( {\frac{1}{x}} \right) \leq x^2 \cr & \mathop {\lim }\limits_{x \to 0} - x^2 = 0\,\,en\,\,\mathop {\lim }\limits_{x \to 0} x^2 = 0\, \cr & \mathop {\lim }\limits_{x \to 0} x^2 \sin \left( {\frac{1}{x}} \right) = 0 \cr} $ ©2004-2024 WisFaq
\require{AMSmath}
Gevraagd: $\eqalign{\mathop {\lim }\limits_{x \to 0} x^2 \sin \left( {\frac{1}{x}} \right)}$ $\eqalign{ & \mathop {\lim }\limits_{x \to 0} x^2 \sin \left( {\frac{1}{x}} \right) \cr & - x^2 \leq x^2 \sin \left( {\frac{1}{x}} \right) \leq x^2 \cr & \mathop {\lim }\limits_{x \to 0} - x^2 = 0\,\,en\,\,\mathop {\lim }\limits_{x \to 0} x^2 = 0\, \cr & \mathop {\lim }\limits_{x \to 0} x^2 \sin \left( {\frac{1}{x}} \right) = 0 \cr} $
Gevraagd: $\eqalign{\mathop {\lim }\limits_{x \to 0} x^2 \sin \left( {\frac{1}{x}} \right)}$
$\eqalign{ & \mathop {\lim }\limits_{x \to 0} x^2 \sin \left( {\frac{1}{x}} \right) \cr & - x^2 \leq x^2 \sin \left( {\frac{1}{x}} \right) \leq x^2 \cr & \mathop {\lim }\limits_{x \to 0} - x^2 = 0\,\,en\,\,\mathop {\lim }\limits_{x \to 0} x^2 = 0\, \cr & \mathop {\lim }\limits_{x \to 0} x^2 \sin \left( {\frac{1}{x}} \right) = 0 \cr} $
©2004-2024 WisFaq