Prove that a lim does not exist
- Thread starter spinos
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topsquark
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Since [math]\dfrac{sin(x)~ sin \left ( \dfrac{1}{x} \right ) }{x}[/math] is continuous for [math]x \neq 0[/math] we can write
[math]\lim_{x \to 0} \dfrac{sin(x)~ sin \left ( \dfrac{1}{x} \right ) }{x} = \left [ \lim_{x \to 0} \dfrac{sin(x)}{x} \right ] \cdot \left [ \lim_{x \to 0} sin \left ( \dfrac{1}{x} \right ) \right ][/math]
The limit of the first factor exists. Can you show that the second factor does not have a limit?
-Dan
[math]\lim_{x \to 0} \dfrac{sin(x)~ sin \left ( \dfrac{1}{x} \right ) }{x} = \left [ \lim_{x \to 0} \dfrac{sin(x)}{x} \right ] \cdot \left [ \lim_{x \to 0} sin \left ( \dfrac{1}{x} \right ) \right ][/math]
The limit of the first factor exists. Can you show that the second factor does not have a limit?
-Dan