AP Calculus AB and BC

Limit of x^2 sin(1/x) at Infinity Is Infinite

Q: What is the limit of x^2 sin(1/x) as x approaches infinity?

It is \infty. The function grows without bound, at roughly the same rate as x.

Q: Why is the answer 0 at x = 0 but infinite at infinity?

At 0 the factor x^{2} crushes a bounded oscillation. At infinity \sin\left(\frac{1}{x}\right) is close to \frac{1}{x}, so it cancels one power of x and one power survives.

Q: Does an infinite limit count as existing?

No finite value exists. Writing \infty is a description of how the limit fails, and on the exam it is the expected answer.

As x approaches infinity, x squared times the sine of 1 over x grows without bound, so the limit is infinite. Write the product as x times the bracket x sin of 1 over x. The bracket tends to 1, so the whole thing grows the way x does.

\lim_{x \to \infty} x^{2}\sin\left(\frac{1}{x}\right) = \infty

Settled by peeling off the standard limit, leaving one factor of x.

Split off the part you already know

x^{2}\sin\left(\frac{1}{x}\right) = x\cdot\left[x\sin\left(\frac{1}{x}\right)\right]

Put $t = \frac{1}{x}$ . As $x \to \infty$ , $t \to 0^{+}$ , and the bracket becomes $\frac{\sin t}{t}$ .

\lim_{x \to \infty} x\sin\left(\frac{1}{x}\right) = \lim_{t \to 0^{+}}\frac{\sin t}{t} = 1

So for large $x$ the function is close to $x$ itself, and $x$ has no ceiling. The limit is infinite.

The same formula behaves oppositely at 0

-x^{2} \le x^{2}\sin\left(\frac{1}{x}\right) \le x^{2}

As $x \to 0$ those outer bounds both go to $0$ , so the squeeze theorem gives $0$ . As $x \to \infty$ the same bounds say only that the function sits between $-x^{2}$ and $x^{2}$ , which rules out nothing.

Which factor is the wild one

Near 0 the sine oscillates and the square tames it. Far out the argument 1 over x is tiny, so the sine is nearly 1 over x and cancels exactly one power of x. Same expression, opposite ends, opposite behaviour.

The mistakes students make

Every one of these comes from importing an argument that belongs at the other end of the axis.

Carrying the squeeze result from $x \to 0$ across to infinity and giving $0$ . The bound $x^{2}$ is useless when $x^{2}$ is enormous.
Stopping once $x\sin\left(\frac{1}{x}\right) \to 1$ and reporting $1$ , which throws away the extra factor of $x$ standing in front of the bracket.
Saying the limit does not exist because sine oscillates, and stopping there. The argument $\frac{1}{x}$ tends to $0$ , so this sine settles down rather than swinging. The values increase without bound, and $\infty$ is the answer that records how the limit fails.

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

What is the limit of x^2 sin(1/x) as x approaches infinity?

It is $\infty$ . The function grows without bound, at roughly the same rate as $x$ .

Why is the answer 0 at x = 0 but infinite at infinity?

At $0$ the factor $x^{2}$ crushes a bounded oscillation. At infinity $\sin\left(\frac{1}{x}\right)$ is close to $\frac{1}{x}$ , so it cancels one power of $x$ and one power survives.

Does an infinite limit count as existing?

No finite value exists. Writing $\infty$ is a description of how the limit fails, and on the exam it is the expected answer.