Back to the on-screen lesson ·
Trapping a limit between two that agree, the fundamental trigonometric limit, and why calculus uses radians.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to apply the squeeze theorem with both of its hypotheses checked, recognise the bounded-times-vanishing pattern and handle it without needing the oscillating factor to have a limit, evaluate the standard trigonometric limits by matching the angle inside the sine to the denominator, derive the cosine limit from the sine one, and say why the limit $\sin u / u \to 1$ is the reason calculus is done in radians.
The squeeze theorem. If $g(x) \le f(x) \le h(x)$ for all $x$ near $a$ (except possibly at $a$), and $$\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L,$$ then $\lim_{x \to a} f(x) = L$.
Two hypotheses, and both are load-bearing. The inequality need only hold near $a$ — a bound that fails far away is no obstacle. And the two outer limits must be equal: bounds that converge to different values trap $f$ in an interval and say nothing whatever about a limit, since a function can be bounded and still oscillate for ever.
What makes the theorem indispensable is that $f$ is never examined. It is not factored, not evaluated, not required to have any structure at all — it only has to sit between two functions that agree in the limit. So it settles limits that no amount of algebra can touch, and it is the only tool in this course that does.
The fundamental trigonometric limit. $$\lim_{u \to 0} \frac{\sin u}{u} = 1.$$ It is proved by squeezing, from the geometric inequality $\cos u \le \dfrac{\sin u}{u} \le 1$ for $0 < |u| < \pi/2$, and it is the reason the derivative of $\sin$ is $\cos$ rather than a constant times $\cos$ — which in turn is why calculus is done in radians.
Its companion is $$\lim_{u \to 0} \frac{1 - \cos u}{u^2} = \frac12,$$ got from the first by multiplying above and below by $1 + \cos u$.
Another way: picture
Two curves closing on the same height at $a$, with a third trapped between them. Whatever the middle one does — wobble, oscillate, jump about — it is squeezed into the same height, because there is nowhere else left for it to be. Now separate the two outer curves so they close on different heights: the middle one has a whole interval to roam in, and the picture stops telling you anything.
Another way: steps
Take $0 < u < \pi/2$ and draw the unit circle. Compare three areas: the triangle with vertices at the origin, $(1,0)$ and $(\cos u, \sin u)$; the circular sector of angle $u$; and the right triangle with the tangent as its far side. Their areas are $$\tfrac12 \sin u \le \tfrac12 u \le \tfrac12 \tan u.$$
Dividing by $\tfrac12 \sin u$ (positive) and inverting: $$\cos u \le \frac{\sin u}{u} \le 1.$$ Both ends tend to $1$, and the squeeze does the rest. The same inequality holds for $u < 0$ because both sides are even.
The sector's area is $\tfrac12 u$ only when $u$ is in radians. In degrees it would be $\tfrac{\pi u}{360}$, the limit would be $\tfrac{\pi}{180}$, and that constant would then appear in the derivative of $\sin$, in every trigonometric integral and in every Taylor series after it. Radians are not a convention chosen for elegance; they are the unit that makes this limit $1$, and the rest of calculus is built on that.
Bounding with the wrong sign. Multiplying $-1 \le \sin\theta \le 1$ by $x$ gives a valid sandwich only when $x \ge 0$. Use $-|x| \le x\sin\theta \le |x|$ and the sign takes care of itself.
Mismatching the angle. $\dfrac{\sin 3x}{x}$ is not $1$. The rule is $\dfrac{\sin u}{u}$ with the same $u$ top and bottom; fix the mismatch by multiplying and dividing, and take the constant outside.
Using the squeeze with unequal bounds. Two bounds with different limits are an interval, not a limit.
Applying $\sin u / u$ away from $0$. The limit is a statement about $u \to 0$. At $u \to \pi$ the quotient tends to $0$, and substitution settles it with no theorem at all.
Forgetting the bound need only hold near $a$. $\sin x \le x$ holds for $x \ge 0$ and fails for $x < 0$ — near $0$ from the right, that is enough.
"$\sin(1/x)$ is between $-1$ and $1$, so it must settle down somewhere" is the intuition the squeeze theorem is most often misused to support. It does not settle anywhere: near $0$ it takes every value in $[-1, 1]$ infinitely often, and it has no limit at all.
What the squeeze theorem adds is not boundedness but a closing gap. $x^2\sin(1/x)$ has a limit because the interval it is confined to shrinks to a point; $\sin(1/x)$ does not, because its interval never shrinks. When you use the theorem, the thing to check is not that the bounds exist but that they meet.
$\lim_{x \to 0} x^2 \cos\!\left(\dfrac{1}{x^2}\right)$. The cosine factor has no limit: as $x \to 0$ its argument runs to infinity and it cycles for ever.
So the product law is unavailable.
But $\left|\cos\left(1/x^2\right)\right| \le 1$, so $-x^2 \le x^2\cos(1/x^2) \le x^2$ — and $x^2 \ge 0$, so no sign reversal to worry about.
Bound, then multiply.
Both outer limits are $0$, so the limit is $0$. The oscillation is real and unending; it is simply confined to a gap that closes.
$\dfrac{1 - \cos u}{u^2}$: multiply above and below by $1 + \cos u$, which is non-zero near $0$.
The conjugate trick again.
The numerator becomes $1 - \cos^2 u = \sin^2 u$, so the expression is $\dfrac{\sin^2 u}{u^2(1 + \cos u)} = \left(\dfrac{\sin u}{u}\right)^2 \cdot \dfrac{1}{1 + \cos u}$.
Now it is built from things with known limits.
The first factor tends to $1^2 = 1$ and the second to $\tfrac{1}{2}$, so the limit is $\tfrac12$ — by the product law, which is now permitted because both pieces have limits.
$\tan 4x = \dfrac{\sin 4x}{\cos 4x}$, so the expression is $\dfrac{\sin 4x}{x} \cdot \dfrac{1}{\cos 4x}$.
Split off the cosine.
The first factor is $4 \cdot \dfrac{\sin 4x}{4x} \to 4$; the second tends to $\dfrac{1}{1} = 1$.
So the limit is $4$.
Match each standard form to its limit conclusion.
| Limit 1 | Limit 0 | Limit 5 | Radians | |
|---|---|---|---|---|
| $\sin x/x$ as $x\to0$ | ||||
| $(1-\cos x)/x$ as $x\to0$ | ||||
| $\sin(5x)/x$ as $x\to0$ | ||||
| The angle unit for these standard limits |
Match each limit observation to the conclusion it supports.
| The middle expression is eligible for squeezing | The squeezed limit is 0 | Its limit is 1 | It is an indeterminate form, not direct substitution | |
|---|---|---|---|---|
| $-x^2\leq x^2\sin(1/x)\leq x^2$ | ||||
| Both $-x^2$ and $x^2$ approach 0 | ||||
| $\sin(3x)/(3x)$ | ||||
| $\sin x/x$ at $x=0$ |
Match each nearby bound to the limit conclusion the squeeze theorem permits.
| Conclude $\lim_{x\to0}f(x)=2$ | No value is forced because the outer limits differ | Conclude $g(x)h(x)\to0$ | Conclude $\lim_{x\to0}\sin x/x=1$ | |
|---|---|---|---|---|
| $2-x^2\leq f(x)\leq2+x^2$ near 0 | ||||
| $0\leq f(x)\leq1$ near 0 | ||||
| $|g(x)|\leq5$ and $h(x)\to0$ | ||||
| $\cos x\leq\sin x/x\leq1$ near 0 |
Find $\displaystyle\lim_{x \to 0} \dfrac{1 - \cos x}{x}$, to four decimal places where it is not a whole number.
Answer:
Find $\displaystyle\lim_{x \to 0} \frac{\sin(3x)}{9x}$.
Answer:
Find $\displaystyle\lim_{x \to 0} \frac{\sin(9x)}{\sin(8x)}$.
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Find $\displaystyle\lim_{x \to 0} x^{6} \sin\!\left(\frac{1}{x}\right)$, or give $0$ if it is zero.
Answer:
You can use the squeeze theorem, and evaluate limits that reduce to $\sin u / u$. Say in your own words why a bounded function need not have a limit, and what the theorem adds to boundedness. Next: continuity, and the two existence theorems it buys.
8. Your turn: $\lim_{x \to 0} \dfrac{\tan 4x}{x}$, step 3