AP Calculus AB and BC

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

Q: Does the same trick work for tangent?

Yes. \frac{\tan^{2}x}{x^{2}} \to 1 for the same reason, since \frac{\tan x}{x} \to 1.

The limit of sine squared x over x squared as x approaches zero is one. The whole expression is the square of sine x over x, and since that tends to one, so does its square.

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

Settled by the square of the special trigonometric limit.

Recognise the square

\frac{\sin^{2}x}{x^{2}} = \left(\frac{\sin x}{x}\right)^{2} \longrightarrow 1^{2} = 1

The power law for limits applies because the inner limit exists, so the limit of the square is the square of the limit. No further work is needed.

Spotting that the expression is a perfect square is the whole problem. Written as $\frac{\sin^{2}x}{x^{2}}$ it can look like it needs L'Hopital twice.

The pattern to carry forward

Any power works the same way: $\frac{\sin^{n}x}{x^{n}} \to 1$ for every positive $n$ , since it is the standard limit raised to that power.

Mismatched powers are the interesting cases. With more sine than $x$ the quotient vanishes, and with more $x$ than sine it is unbounded. Counting powers on each side answers all three at once.

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

Do I need L'Hopital's rule here?

No, and it would take two applications. Recognising the square reduces the problem to a limit you already know.

Does the same trick work for tangent?

Yes. $\frac{\tan^{2}x}{x^{2}} \to 1$ for the same reason, since $\frac{\tan x}{x} \to 1$ .