AP Calculus AB and BC

Limit of x sin(1/x) as x Approaches 0 Is 0

Q: How can the limit exist when sin(1/x) has no limit?

A limit describes the product, not each factor separately. Every output obeys \left|x\sin\frac{1}{x}\right| \le |x|, so once |x| < 0.001 every output is within 0.001 of 0, oscillation and all. The definition of the limit asks for nothing more than that.

Q: Is the function continuous at x = 0?

Not as written, since x = 0 is outside the domain. Because the limit exists, the gap is a removable discontinuity: the piecewise function with value 0 at the origin is continuous everywhere. It is still not differentiable there, since the difference quotient \frac{h\sin\frac{1}{h}}{h} = \sin\frac{1}{h} has no limit.

Q: Does the same argument work for x sin(1/x) as x approaches infinity?

No, and the answer changes. Put u = \frac{1}{x}, so the expression becomes \frac{\sin u}{u} with u \to 0, giving 1. The squeeze bound |x| is useless out there because it grows instead of shrinking.

The limit of x sin(1/x) as x approaches 0 is 0. Substitution fails because sin(1/x) has no limit at 0: it oscillates faster and faster as x shrinks. The squeeze theorem settles it, since the function stays between minus the absolute value of x and the absolute value of x, and both bounds go to 0.

\lim_{x \to 0} x\sin\frac{1}{x} = 0

Settled by the squeeze theorem.

Trapping the oscillation between two lines

Only one fact about the sine factor is needed, and it holds no matter how wild the input $\frac{1}{x}$ gets: sine never leaves the interval from $-1$ to $1$ .

-1 \le \sin\frac{1}{x} \le 1 \quad (x \neq 0)

Multiplying an inequality by $x$ would flip it whenever $x$ is negative, so multiply by $|x|$ instead. Since $\left|x\sin\frac{1}{x}\right| = |x|\left|\sin\frac{1}{x}\right| \le |x|$ , the function lives inside a wedge.

-|x| \le x\sin\frac{1}{x} \le |x|

Both outer functions are continuous and hit 0 at the origin, so they close on the same value.

\lim_{x \to 0}\left(-|x|\right) = 0 = \lim_{x \to 0}|x|

With the upper and lower bounds pinned to 0, whatever sits between them is pinned there too. The graph shows the same thing: the oscillations are packed into the wedge between $y = x$ and $y = -x$ , and that wedge pinches shut at the origin.

Bounded times vanishing gives 0

A factor that stays inside fixed bounds, multiplied by something heading to 0, gives 0 whether or not the bounded factor has a limit of its own. That is why $x\cos\frac{1}{x}$ , $x\sin\frac{1}{x^{2}}$ , and $\sqrt{x}\sin\frac{1}{x}$ all go to 0 as well.

Why substitution and the product law fail

Substituting $x = 0$ asks for $\sin\frac{1}{0}$ , which is not a number, so the function has no value at 0 to read off. The honest description of what substitution returns is $0 \cdot (\text{no limit})$ , and that is a blocker rather than one of the named indeterminate forms.

The product law is the next reflex, and it is unavailable for the same reason. Splitting $\lim (fg)$ into $\lim f \cdot \lim g$ requires both limits to exist, and the second one does not.

\lim_{x \to 0}\sin\frac{1}{x} \quad \text{does not exist}

As $x \to 0$ the input $\frac{1}{x}$ runs through every large value, so the sine factor equals 1 at $x = \frac{2}{(4k+1)\pi}$ and $-1$ at $x = \frac{2}{(4k+3)\pi}$ , infinitely many times inside any interval around 0. It never settles.

L'Hopital's rule is also out. Its hypotheses need $\frac{0}{0}$ or $\frac{\infty}{\infty}$ , and this is a product with a factor that has no limit at all. The squeeze theorem is the tool built for exactly this situation, because it never asks the oscillating factor to converge.

The mistakes students make

Answering "does not exist" because $\sin\frac{1}{x}$ does not. The oscillation never stops, but its amplitude is being crushed by the $x$ in front, and 0 is a perfectly good limit.
Writing the squeeze as $-x \le x\sin\frac{1}{x} \le x$ . That is false for $x < 0$ , where the outer expressions swap order. Use $|x|$ , or handle the two sides separately.
Bounding the wrong factor, as in $-x \le \sin\frac{1}{x} \le x$ . Sine is the bounded piece; $x$ is the piece going to 0.
Confusing this with $\frac{\sin x}{x}$ , whose limit at 0 is 1. Different function, different technique, different answer.
Calling the function continuous at 0. As written it is undefined there, so the discontinuity is removable: defining the value at 0 to be 0 patches it.

Not sure which technique a limit wants?

The Limit Method Chooser walks the decision from direct substitution through factoring, the conjugate, and L'Hopital, and says why each one applies or fails.

Frequently asked questions

How can the limit exist when sin(1/x) has no limit?

A limit describes the product, not each factor separately. Every output obeys $\left|x\sin\frac{1}{x}\right| \le |x|$ , so once $|x| < 0.001$ every output is within $0.001$ of 0, oscillation and all. The definition of the limit asks for nothing more than that.

Is the function continuous at x = 0?

Not as written, since $x = 0$ is outside the domain. Because the limit exists, the gap is a removable discontinuity: the piecewise function with value 0 at the origin is continuous everywhere. It is still not differentiable there, since the difference quotient $\frac{h\sin\frac{1}{h}}{h} = \sin\frac{1}{h}$ has no limit.

Does the same argument work for x sin(1/x) as x approaches infinity?

No, and the answer changes. Put $u = \frac{1}{x}$ , so the expression becomes $\frac{\sin u}{u}$ with $u \to 0$ , giving 1. The squeeze bound $|x|$ is useless out there because it grows instead of shrinking.