Exercise · ZW-LIMIT-OF-A-FUNCTION-017 · ★★☆ Intermediate
Which pair of sequences tending to \(0\) proves that \(\sin(1/x)\) has no limit at \(0\)?
Choose one option.
Show answer
- C. \(x_n=\dfrac1{n\pi}\) and \(y_n=\dfrac1{\pi/2+2n\pi}\)
Exercise · ZW-LIMIT-OF-A-FUNCTION-017 · ★★☆ Intermediate
Choose one option.
Along \(x_n\) the value is \(\sin(n\pi)=0\) and along \(y_n\) it is \(\sin(\pi/2+2n\pi)=1\): two different limits. In the other pairs both sequences give the same value (\(0\) or \(1\)), so nothing follows.
The limit of \(f\) as \(x\) approaches \(a\) is \(L\) if \(f(x)\) gets as close to \(L\) as we like when \(x\) is taken sufficiently close to \(a\), regardless of what happens at \(x=a\).