AP Calculus AB and BC

Squeeze Theorem vs L'Hopital's Rule

Q: Do I use L'Hopital's rule or the squeeze theorem for x sin(1/x)?

The squeeze theorem. Since \lvert x \sin(1/x) \rvert \le \lvert x \rvert, the expression is trapped between -\lvert x \rvert and \lvert x \rvert, so the limit at 0 is 0. L'Hopital's rule has no role here, because a bounded factor times something going to zero is not an indeterminate quotient.

Q: Does the squeeze theorem work for limits at infinity?

Yes, with the same wording. For x > 0 the bounds -\frac{1}{x} \le \frac{\sin x}{x} \le \frac{1}{x} both go to zero, so \lim_{x \to \infty} \frac{\sin x}{x} = 0. L'Hopital has no entry point here, because the numerator is bounded and has no limit, so the form is not infinity over infinity.

L'Hopital's rule applies only to a differentiable quotient in the form 0 over 0 or infinity over infinity. The squeeze theorem needs only upper and lower bounds that share a limit, so it handles oscillating limits like x squared times sine of 1 over x, where L'Hopital does not apply.

Squeeze theorem

Use when: A bounded factor is multiplied by something heading to zero, so the whole expression can be trapped between two functions that share a limit.

L'Hopital's rule

Use when: Substitution returns zero over zero or infinity over infinity, both pieces are differentiable near the point, and their derivatives are tidier than they are.

Side by side

	Squeeze theorem	L'Hopital's rule
What you must supply	Functions with $g \le f \le h$ and $\lim g = \lim h$	A quotient whose substitution gives $\frac{0}{0}$ or $\frac{\infty}{\infty}$
Differentiability	Not needed, and $f$ need not be defined at the point	Required of the numerator and the denominator
Bounded oscillation	Handles it directly, that is what the theorem is for	Usually makes it worse, because the derivative oscillates too
Fails when	No pair of bounds with a common limit is available	The new limit $\lim f'(x)/g'(x)$ does not exist, which settles nothing
Standard case	$\lim_{x \to 0} x^{2}\sin(1/x) = 0$	$\lim_{x \to \infty} \frac{\ln x}{x} = 0$

The squeeze theorem asks for company. If $g(x) \le f(x) \le h(x)$ for every $x$ near $a$ , and $g$ and $h$ both approach the same value $L$ there, then $f$ has nowhere else to go and its limit is $L$ as well. Nothing in that argument mentions derivatives, and $f$ does not even have to be defined at $a$ . Since $\lvert \sin(1/x) \rvert \le 1$ for every $x \ne 0$ , multiplying through by $x^{2}$ produces bounds that both collapse to zero.

-x^{2} \le x^{2}\sin\left(\frac{1}{x}\right) \le x^{2} \qquad \Longrightarrow \qquad \lim_{x \to 0} x^{2}\sin\left(\frac{1}{x}\right) = 0

L'Hopital's rule asks for a quotient in an indeterminate form. That expression is not one. Written as $\sin(1/x) \div x^{-2}$ the denominator grows without bound but the numerator has no limit at all, so the form is neither $\frac{0}{0}$ nor $\frac{\infty}{\infty}$ , and every other rearrangement either leaves a factor with no limit or puts zeros in the denominator at points arbitrarily close to $0$ . The rule has no entry point.

A failed L'Hopital is not a proof that the limit is missing

The rule runs one way only: if $\lim f'(x)/g'(x)$ exists then the original limit matches it, and if that new limit does not exist you have learned nothing. Students who differentiate an oscillating quotient, watch the result swing forever, and then write that the original limit does not exist have drawn a conclusion the theorem never offers. Take $\frac{x + \sin x}{x}$ as $x \to \infty$ : the derivatives give $1 + \cos x$ , which never settles, while the bounds $\frac{x - 1}{x} \le \frac{x + \sin x}{x} \le \frac{x + 1}{x}$ give the answer $1$ immediately.

Frequently asked questions

Do I use L'Hopital's rule or the squeeze theorem for x sin(1/x)?

The squeeze theorem. Since $\lvert x \sin(1/x) \rvert \le \lvert x \rvert$ , the expression is trapped between $-\lvert x \rvert$ and $\lvert x \rvert$ , so the limit at $0$ is $0$ . L'Hopital's rule has no role here, because a bounded factor times something going to zero is not an indeterminate quotient.

Does the squeeze theorem work for limits at infinity?

Yes, with the same wording. For $x > 0$ the bounds $-\frac{1}{x} \le \frac{\sin x}{x} \le \frac{1}{x}$ both go to zero, so $\lim_{x \to \infty} \frac{\sin x}{x} = 0$ . L'Hopital has no entry point here, because the numerator is bounded and has no limit, so the form is not infinity over infinity.

If L'Hopital's rule gives a limit that does not exist, has the original limit failed?

The original limit has not failed. That outcome means only that the rule is inconclusive and you have to find another route, and the limit is often a perfectly ordinary number that bounds will reveal in one line.

In the CED: Unit 1: Limits and Continuity, Unit 4: Contextual Applications