AP Calculus AB and BC

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

Q: What is the limit of sin(x^2) / x as x approaches 0?

It is 0. Rewrite as x\cdot\frac{\sin(x^{2})}{x^{2}}: the quotient tends to 1 and the leading x tends to 0.

Q: Why is it not 1?

Because the inner angle x^{2} does not match the denominator x. Fixing the match leaves an extra factor of x behind.

Q: How is sin(x^2) different from sin^2 x?

\sin(x^{2}) squares the input before taking sine; \sin^{2}x takes sine and squares the result.

The limit of sin of x squared, divided by x, as x approaches 0 is 0. The special trigonometric limit needs the inner angle to match the denominator, and here it is x squared over x. Writing it as x times sin(x squared) over x squared leaves a stray factor of x, which goes to 0.

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

Settled by matching the inner angle, leaving a stray factor.

Manufacturing the match

The inner angle is $x^{2}$ , so the special limit wants $x^{2}$ underneath. Multiply and divide to get it, and see what is left over.

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

The quotient tends to $1$ by the special limit with $u = x^{2}$ , but the factor of $x$ in front does not disappear, and it is heading to $0$ .

\lim_{x \to 0}\frac{\sin\left(x^{2}\right)}{x} = 0 \cdot 1 = 0

Do not confuse it with sin^2 x over x

That one is also 0, but for a different reason: sin squared x over x equals sin x times sin x over x, which is 0 times 1. Both give 0, and neither is the special limit on its own.

The L'Hopital check

The form is $\frac{0}{0}$ , and the chain rule handles the numerator.

\lim_{x \to 0}\frac{\sin\left(x^{2}\right)}{x} \;\overset{\text{H}}{=}\; \lim_{x \to 0}\frac{2x\cos\left(x^{2}\right)}{1} = 0

The surviving factor of $x$ in the algebraic method shows up here as the $2x$ from the chain rule. Both routes are recording the same fact: the numerator vanishes faster than the denominator.

The mistakes students make

Answering $1$ on sight. The special limit requires the inner angle and the denominator to be identical, and $x^{2}$ is not $x$ .
Reading $\sin\left(x^{2}\right)$ as $\sin^{2}x$ . They are different functions: one squares the input, the other squares the output.
Concluding the limit does not exist because substitution fails. $\frac{0}{0}$ is a signal to do more work, not a verdict.

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 sin(x^2) / x as x approaches 0?

It is $0$ . Rewrite as $x\cdot\frac{\sin(x^{2})}{x^{2}}$ : the quotient tends to $1$ and the leading $x$ tends to $0$ .

Why is it not 1?

Because the inner angle $x^{2}$ does not match the denominator $x$ . Fixing the match leaves an extra factor of $x$ behind.

How is sin(x^2) different from sin^2 x?

$\sin(x^{2})$ squares the input before taking sine; $\sin^{2}x$ takes sine and squares the result.