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 \(y_n=\dfrac1{\pi/2+2n\pi}\)

Explanation

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.

Study Calculus in ZWMath →

Review the concept: Limit of a function

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 \(x=a\).

\[\lim_{x\to a} f(x)=L\]

More practice problems: limit of a function

← ZW-LIMIT-OF-A-FUNCTION-016 ZW-LIMIT-OF-A-FUNCTION-018 →