AP Calculus AB and BC

Limit of sin x / sqrt x at 0 From the Right Is 0

Q: What is the limit of sin x / sqrt x as x approaches 0 from the right?

It is 0, since \frac{\sin x}{\sqrt{x}} = \frac{\sin x}{x}\cdot\sqrt{x} and the second factor drives the product to zero.

Q: Why can the limit only be taken from the right?

\sqrt{x} is undefined for x < 0, so the function has no graph on the left of the origin and only the right-hand limit is meaningful.

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

Yes. Differentiating gives \frac{\cos x}{\frac{1}{2\sqrt{x}}} = 2\sqrt{x}\cos x, which tends to 0 as x \to 0^{+}.

The limit of sin x over the square root of x as x approaches 0 from the right is 0. Rewrite it as sin x over x, times the square root of x: the first factor tends to 1 and the second to 0, so the product tends to 0.

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

Settled by splitting off the special trigonometric limit.

Split off the standard limit

The quotient is a $\frac{0}{0}$ form. Insert the denominator the standard limit wants, then repair the expression with the factor that is left over.

\frac{\sin x}{\sqrt{x}} = \frac{\sin x}{x}\cdot\frac{x}{\sqrt{x}} = \frac{\sin x}{x}\cdot\sqrt{x}

The step $\frac{x}{\sqrt{x}} = \sqrt{x}$ holds for $x > 0$ , which is exactly the region a right-hand limit looks at. Now the two factors have limits $1$ and $0$ .

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

Why only one side is available

For $x < 0$ the expression $\sqrt{x}$ is not a real number, so the function has no values to the left of $0$ . There is nothing to approach along, and the two-sided limit does not exist for want of a left side.

On the right, the working above says the function behaves like $\sqrt{x}$ near the origin, because $\sin x$ behaves like $x$ . That is why the graph leaves the origin steeply, with a vertical tangent, and still arrives at height $0$ .

Compare the two rates

The numerator and denominator both head to zero, so the winner is whichever gets there faster. Sine behaves like x, and x reaches zero faster than the square root of x does, so the quotient is dragged down to zero.

The mistakes students make

Pattern matching too early produces the first two answers below, while the third ignores the domain.

Answering $1$ by reading the quotient as the standard limit. The denominator is $\sqrt{x}$ , not $x$ , and the pattern only applies when the two match.
Answering $\infty$ because the denominator tends to $0$ . The numerator tends to $0$ faster, so the quotient collapses instead of blowing up.
Reporting the two-sided limit as $0$ as well. The function is undefined for negative $x$ , so the left-hand limit does not exist and neither does the two-sided one.

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 / sqrt x as x approaches 0 from the right?

It is $0$ , since $\frac{\sin x}{\sqrt{x}} = \frac{\sin x}{x}\cdot\sqrt{x}$ and the second factor drives the product to zero.

Why can the limit only be taken from the right?

$\sqrt{x}$ is undefined for $x < 0$ , so the function has no graph on the left of the origin and only the right-hand limit is meaningful.

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

Yes. Differentiating gives $\frac{\cos x}{\frac{1}{2\sqrt{x}}} = 2\sqrt{x}\cos x$ , which tends to $0$ as $x \to 0^{+}$ .