← Back to learning objective

Existence of the limit of sin(1/x) as x approaches 0

Consider two related functions defined for x≠0x \neq 0: d(x)=xsin⁡(1/x)d(x) = x\sin(1/x) and e(x)=sin⁡(1/x)e(x) = \sin(1/x). Both involve the factor sin⁡(1/x)\sin(1/x), which oscillates faster and faster as x→0x \to 0, but the two functions behave very differently near x=0x = 0.

The graph below shows e(x)=sin⁡(1/x)e(x) = \sin(1/x) near x=0x = 0.

Existence of the limit of sin(1/x) as x approaches 0 graph

As xx gets closer and closer to 00 (from either side), does e(x)=sin⁡(1/x)e(x) = \sin(1/x) approach a single number LL? Based on the behavior shown, does lim⁡x→0sin⁡(1/x)\displaystyle\lim_{x\to 0} \sin(1/x) exist? If it does not exist, state which of the standard reasons (unequal one-sided limits, unbounded growth, or oscillation without settling) applies.

Hint

Think about what happens to 1/x1/x as x→0x \to 0: it takes on larger and larger magnitude, so sin⁡(1/x)\sin(1/x) keeps cycling through all values between −1-1 and 11 infinitely often, no matter how close xx gets to 00. Ask yourself whether the outputs ever settle down near a single value.

Step-by-step solution

As x→0x \to 0, the quantity 1/x1/x grows without bound in magnitude, so it sweeps repeatedly through every real number value, causing sin⁡(1/x)\sin(1/x) to oscillate between −1-1 and 11 infinitely many times no matter how small ∣x∣|x| is. Unlike d(x)=xsin⁡(1/x)d(x) = x\sin(1/x), there is no shrinking factor multiplying the oscillation here, so the oscillation is never damped.

Because e(x)e(x) keeps cycling through all values in [−1,1][-1,1] arbitrarily close to x=0x=0, it never settles near a single value LL as x→0x \to 0. This is the oscillation-without-settling case, so lim⁡x→0sin⁡(1/x)\lim_{x\to 0}\sin(1/x) does not exist.

Related problems

Practice similar Calculus problems interactively.

Start Closing the Gap