\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} ©2004-2025 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}
©2004-2025 WisFaq