AP Calculus AB and BC

Limit of x^2 sin(1/x) as x Approaches 0 Is 0

Q: What is the limit of x^3 sin(1/x) or x^2 cos(1/x)?

Both are 0, by the identical argument. Any x^{n} with n \ge 1 times any factor confined between two constants goes to 0 at the origin. Only the bounding function changes: |x|^{n} replaces x^{2}.

The limit of x squared times sin(1/x) as x approaches 0 is 0. Substitution fails because sin(1/x) oscillates without settling. Since x squared is never negative, the squeeze is clean: the function sits between negative x squared and x squared, and both bounds go to 0.

\lim_{x \to 0} x^2\sin\frac{1}{x} = 0

Settled by the squeeze theorem.

Squeezing between the two parabolas

Start from the one bound sine always obeys, then multiply through by $x^{2}$ . Because $x^{2} \ge 0$ for every $x$ , the inequality keeps its direction and no absolute values are needed.

-1 \le \sin\frac{1}{x} \le 1 \;\Longrightarrow\; -x^{2} \le x^{2}\sin\frac{1}{x} \le x^{2}

\lim_{x \to 0}\left(-x^{2}\right) = 0 = \lim_{x \to 0}x^{2}

The two bounds meet at 0, so the function trapped between them is forced to 0. Graphically the oscillations are held inside the narrowing throat between $y = x^{2}$ and $y = -x^{2}$ , touching each parabola infinitely often on the way in.

Why the squared version is the tidier one

With $x\sin\frac{1}{x}$ you have to switch to $\pm|x|$ , since multiplying an inequality by a negative $x$ reverses it. Squaring removes the sign question entirely, which is why textbooks reach for $x^{2}\sin\frac{1}{x}$ when they want the squeeze theorem with no side conditions.

Why direct substitution fails

At $x = 0$ the factor $\sin\frac{1}{0}$ is undefined, so there is nothing to substitute into. Nor can the limit be split into $\lim x^{2} \cdot \lim \sin\frac{1}{x}$ , because the product law requires both pieces to converge and the sine factor does not.

A table shows what is happening. The sign jumps around, since the sine factor keeps sweeping from $-1$ to 1, while the size collapses at the rate of $x^{2}$ .

$x$	$x^{2}\sin\frac{1}{x}$
0.1	-0.005440
0.01	-0.000051
0.001	0.00000083

Only the magnitudes matter for the squeeze, and they are bounded by $x^{2}$ . The erratic signs are exactly why a numerical table alone is not proof here, and why the inequality argument is the one that closes the case.

Where this function earns its keep

Patch the hole by defining $f(0) = 0$ and the function becomes the standard counterexample in differentiability. Its derivative at the origin comes straight from the difference quotient.

f'(0) = \lim_{h \to 0}\frac{h^{2}\sin\frac{1}{h} - 0}{h} = \lim_{h \to 0}h\sin\frac{1}{h} = 0

That inner limit is the companion squeeze, so $f$ is differentiable at 0. Away from the origin, the product and chain rules give the derivative.

f'(x) = 2x\sin\frac{1}{x} - \cos\frac{1}{x} \quad (x \neq 0)

The first term goes to 0, but $\cos\frac{1}{x}$ oscillates forever, so $\lim_{x \to 0}f'(x)$ does not exist even though $f'(0) = 0$ does. A derivative can exist at a point without being continuous there, which is the whole reason this example is famous.

The mistakes students make

Reporting no limit because the graph never stops wiggling. Bounded oscillation multiplied by something vanishing still vanishes.
Bounding the wrong factor, as in $-x^{2} \le \sin\frac{1}{x} \le x^{2}$ . Sine is the bounded piece and $x^{2}$ is the piece doing the crushing.
Reaching for L'Hopital's rule. The form is not $\frac{0}{0}$ or $\frac{\infty}{\infty}$ , and differentiating produces the same non-converging $\cos\frac{1}{x}$ anyway.
Assuming $f'$ is continuous at 0 because $f'(0)$ exists. This function is the counterexample to that assumption.
Reading the sign changes in a table as evidence of two different one-sided limits. Both sides are squeezed to the same 0.

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 is this different from x sin(1/x)?

Both limits at 0 are 0, and both need the squeeze theorem. The difference shows up one derivative later: with the hole patched, $x\sin\frac{1}{x}$ is continuous at 0 but not differentiable there, while $x^{2}\sin\frac{1}{x}$ is differentiable with $f'(0) = 0$ . The extra factor of $x$ buys the extra smoothness.

Does the squeeze theorem need the function to be defined at 0?

No. Every limit statement ignores the point itself, so the inequalities only have to hold on an interval around 0 with the origin possibly removed. That is why the argument works even though $\sin\frac{1}{0}$ is meaningless.

What is the limit of x^3 sin(1/x) or x^2 cos(1/x)?

Both are 0, by the identical argument. Any $x^{n}$ with $n \ge 1$ times any factor confined between two constants goes to 0 at the origin. Only the bounding function changes: $|x|^{n}$ replaces $x^{2}$ .