AP Calculus AB and BC

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

Q: Does L'Hopital's rule give the same answer?

Yes. The form is \frac{0}{0}, and one pass gives \frac{\cos x}{1+2x}, which is \frac{1}{1} = 1 at the origin. Unit 1 expects the factoring route, so use L'Hopital as a check rather than as the shown work.

Q: What happens if the denominator is x^2 + x^3 instead?

The limit fails to exist. Factoring gives \frac{\sin x}{x^{2}(1+x)} = \frac{1}{x} \cdot \frac{\sin x}{x} \cdot \frac{1}{1+x}, and the \frac{1}{x} sends the right side to +\infty and the left side to -\infty. One extra power of x changes everything.

The limit of sin x/(x + x^2) as x approaches 0 is 1. Direct substitution gives 0 over 0. Factoring the denominator as x times (1 + x) splits the expression into sin x over x, which goes to 1, times 1 over (1 + x), which also goes to 1. The quadratic term is negligible near 0.

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

Settled by the special trigonometric limit.

Factoring to expose the special limit

The denominator is not $x$ , so the standard trig limit does not apply as written. One factoring step fixes that.

\frac{\sin x}{x+x^{2}} = \frac{\sin x}{x(1+x)} = \frac{\sin x}{x} \cdot \frac{1}{1+x}

Each factor now has a limit of its own, so the product law is available: the first is the memorised trig limit, and the second is continuous at 0.

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

Why the quadratic term drops out

Near the origin $x^{2}$ is tiny compared with $x$ . At $x = 0.01$ the denominator is $0.0101$ rather than $0.01$ , a difference of one percent, and the gap shrinks with $x$ . Far from 0 the $x^{2}$ dominates, but a limit at 0 only ever sees the neighbourhood.

Why direct substitution fails

Both parts vanish at the origin, since $\sin 0 = 0$ and $0 + 0^{2} = 0$ .

\frac{\sin 0}{0+0^{2}} = \frac{0}{0}

That form reports a tie and leaves the winner undecided. Here both parts are first order near 0, because $\sin x$ behaves like $x$ and $x + x^{2}$ behaves like $x$ , so the quotient settles at a finite nonzero number.

$x$	$\frac{\sin x}{x+x^{2}}$
-0.1	1.109260
0.1	0.907577
0.01	0.990083
0.001	0.999001

The two sides approach 1 from opposite directions, and the convergence is slow compared with $\frac{\sin x}{x}$ alone, because the $1+x$ factor is what is drifting.

The mistakes students make

Cancelling the $x$ of $\sin x$ against the $x$ in the denominator. In $\sin x$ the $x$ is an argument, not a factor, and nothing about sine can be cancelled away.
Splitting into $\frac{\sin x}{x} + \frac{\sin x}{x^{2}}$ . A sum in a denominator does not split like that: at $x = 0.1$ the true value is about $0.908$ , while that sum is about $10.98$ .
Answering 0 on the grounds that the denominator has more terms. Term count is irrelevant; the lowest-order term is what decides.
Keeping the $1+x$ factor and reporting $\frac{1}{1+x}$ or 0 as the answer, having taken $\frac{\sin x}{x}$ to be 0 rather than 1.
Evaluating with the calculator in degree mode, which turns the answer into roughly $0.0175$ .

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

Does L'Hopital's rule give the same answer?

Yes. The form is $\frac{0}{0}$ , and one pass gives $\frac{\cos x}{1+2x}$ , which is $\frac{1}{1} = 1$ at the origin. Unit 1 expects the factoring route, so use L'Hopital as a check rather than as the shown work.

What happens if the denominator is x^2 + x^3 instead?

The limit fails to exist. Factoring gives $\frac{\sin x}{x^{2}(1+x)} = \frac{1}{x} \cdot \frac{\sin x}{x} \cdot \frac{1}{1+x}$ , and the $\frac{1}{x}$ sends the right side to $+\infty$ and the left side to $-\infty$ . One extra power of $x$ changes everything.

Why does the answer not depend on the x^2 term at all?

Because the limit is decided by the lowest-order behaviour, and $x^{2}$ is a higher order than $x$ . Any denominator of the form $x + cx^{2} + dx^{3} + \cdots$ gives the same limit of 1, since everything past the linear term contributes nothing at the origin.