Using the Squeeze Theorem to Show lim⁡x→0sin⁡(x)x=1begin{aligned}lim_{xto 0}frac{sin(x)}{x} = 1end{aligned} x → 0 lim ​ x sin ( x ) ​ = 1 ​

This was written for my Fall 2023 calculus students in M408C, “Differential and Integral Calculus”.

In one of the recitation classes a while back we needed to either use the following fact:

lim⁡x→0sin⁡(x)x=1.begin{aligned} lim_{xto 0}frac{sin(x)}{x} = 1. end{aligned}x→0lim​xsin(x)​=1.​

At the time, I forgot how to calculate this limit without using l’Hopital’s rule, Taylor Series or the “Small Angle Approximation”. This is a problem, because:

• We hadn’t seen l’Hopital’s rule yet, and worse, it requires knowing that ddxsin⁡(x)=cos⁡(x)dfrac{d}{dx} sin(x) = cos(x)dxd​sin(x)=cos(x). As far as I’m aware, calculating the derivative of sin⁡(x)sin(x)sin(x) requires knowing the above limit, so this method is circular!

• Taylor Series don’t show up until Calculus II and they also require the derivative of sin⁡(x)sin (x)sin(x).

• The small angle approximation says that when xxx is small, sin⁡(x)≈xsin(x) approx xsin(x)≈x. This is quite useful in physics, but is imprecise. To make it precise, you might say that as xxx gets super small, sin⁡(x)=x+ϵsin(x) = x + epsilonsin(x)=x+ϵ where ϵepsilonϵ is some error term, and that ϵepsilonϵ shrinks faster than xxx does as x→0x to 0x→0 (meaning that the percentage error in the approximation sin⁡(x)≈xsin(x) approx xsin(x)≈x goes to zero as x→0xto 0x→0). However, this is a fancy way of saying sin⁡(x)x→1frac{sin(x)}{x} to 1xsin(x)​→1 as x→0xto 0x→0, so this is also circular!

The correct way to calculate this limit, or at least, a correct way to calculate it, is by using the Squeeze Theorem. It’s a wonderful computation tool. It’s doubly embarrassing that I forgot it that day, especially because we used it that same worksheet to find a very similar limit: lim⁡x→0xsin⁡(1/x)lim_{xto 0} xsin(1/x)limx→0​xsin(1/x)! I wrote this post explaining the solution to this problem and delving into a lot of detail about the squeeze theorem, lest you forget it like I did.

Thanks to Catherine Chi for reminding me of this solution and for suggesting we use the areas of triangles to come up with good bounds for the function sin⁡(x)/xsin(x)/xsin(x)/x.

The squeeze theorem

This may be hard to parse, so let’s go over the statement of this theorem in more detail.

• “Let fff, ggg and hhh be real valued functions on some interval containing the real number aaa” means that fff, ggg and hhh are all functions whose domain includes an interval (u,v)⊆R(u,v) subseteq mathbb R(u,v)⊆R and whose codomain is the real numbers Rmathbb RR. That is, the functions you’re used to thinking about. It also names a special number aaa that is between uuu and vvv.

• “Suppose there exists some ϵepsilonϵ such that for all a≠x∈(a−ϵ,a+ϵ)aneq xin (a-epsilon, a+epsilon)a=x∈(a−ϵ,a+ϵ) we have…” is a fancy way of saying “for all points close to aaa but not equal to aaa…”. The number ϵ>0epsilon>0ϵ>0 can be super small, as long as it’s still positive, and thus (a−ϵ,a+ϵ)(a – epsilon, a + epsilon)(a−ϵ,a+ϵ) is a small interval centered at aaa. We call this “a small neighborhood around aaa” informally. This sort of language shows up in the formal definition of a limit, but that’s a story for another time.

• The statement g(x)≤f(x)≤h(x)g(x) leq f(x) leq h(x)g(x)≤f(x)≤h(x) simply means that we want ggg and hhh to be lower and upper bounds of fff respectively. At this point we can see that the last portion of the statement actually gives us more freedom; we don’t need ggg and hhh to bound fffeverywhere, just around aaa. Note that we don’t need ggg and hhh to bound fff at aaa itself; limits don’t care about what happens at the limit point. As per usual we are only interested in what’s happening around aaa.

• “If lim⁡x→ag(x)=lim⁡x→ah(x)=Llim_{xto a} g(x) = lim_{xto a} h(x) = Llimx→a​g(x)=limx→a​h(x)=L…” means, “If the limits of g and h as xxx approaches aaa both exist and are equal…”

• “…then lim⁡x→af(x)=Llim_{xto a} f(x) = Llimx→a​f(x)=L as well,” means “then the limit of fff as xxx approaches aaaexists and is equal to LLL”.

That’s still a big wall of text, and while it’s convincing, perhaps you still don’t feel the truth of the squeeze theorem. Here are two final things that hopefully help you feel it’s truth.

Feeling the truth in your gut

Here’s a slogan for you:

This means that if g(x)≤f(x)≤h(x)g(x) leq f(x) leq h(x)g(x)≤f(x)≤h(x) as above, then lim⁡x→ag(x)≤lim⁡x→af(x)≤lim⁡x→ah(x)lim_{xto a} g(x)leq lim_{xto a} f(x) leq lim_{xto a} h(x)limx→a​g(x)≤limx→a​f(x)≤limx→a​h(x). You have to be careful: if we have strict inequalities g(x)<f(x)<h(x)g(x) < f(x) < h(x)g(x)<f(x)<h(x), then we can only say that lim⁡x→ag(x)≤lim⁡x→af(x)≤lim⁡x→ah(x)lim_{xto a} g(x)leq lim_{xto a} f(x) leq lim_{xto a} h(x)limx→a​g(x)≤limx→a​f(x)≤limx→a​h(x); limits turn strict inequalities into non-strict inequalities. If this fact is giving you pause, draw some pictures (perhaps some example graphs of ggg, fff and hhh) and try to convince yourself that it is true!

If we accept this fact, then in the scenario that lim⁡x→ag(x)=lim⁡x→ah(x)=Llim_{xto a} g(x) = lim_{xto a} h(x) = Llimx→a​g(x)=limx→a​h(x)=L we simply get that

g(x)≤f(x)≤h(x)  ⟹  L≤lim⁡x→af(x)≤L.begin{aligned} g(x) leq f(x) leq h(x) implies L leq lim_{xto a} f(x) leq L. end{aligned}g(x)≤f(x)≤h(x)⟹L≤x→alim​f(x)≤L.​

The only way something can be simultaneously smaller and bigger than something else is if they’re equal! This is one of Terry Tao’s problem solving strategies: break up equalities into inequalities).

Feeling the truth in your gut with a picture

Remember this weird function?

While f(x)=xsin⁡(1/x)f(x) = xsin(1/x)f(x)=xsin(1/x) looks bizarre, it’s bounded above by ∣x∣|x|∣x∣ and below by −∣x∣-|x|−∣x∣, as you can see from the graph. At x=0x = 0x=0, we get ∣x∣=−∣x∣=0|x| = -|x| = 0∣x∣=−∣x∣=0, and so visually, it looks like f(x)f(x)f(x) is being squeezed into 000. Nonetheless, let’s carefully write down a calculation of the limit as x→0xto 0x→0 using the squeeze theorem.

A careful calculation

Notice what happened here: we spent all our work finding upper and lower bounds. Once we had them, the calculation of the limit was immediate.

Takeaway: The squeeze theorem lets you replace the problem of calculating a difficult limit with the problem of finding nice upper and lower bounds.

A solution to the problem

Let’s turn to the problem at hand.

Show the following is true:

lim⁡x→0sin⁡(x)x=1.begin{aligned} lim_{xto 0} frac{sin(x)}{x} = 1. end{aligned}x→0lim​xsin(x)​=1.​

Our strategy is to find functions ggg and hhh which bound sin⁡(x)/xsin(x)/xsin(x)/x near 000 and which have the same limit at 000. Finding functions which bound sin⁡(x)/xsin(x)/xsin(x)/x is easy; the tricky part is ensuring they have the same limit.

Step 1: Convince ourselves the limit exists and is 1.

I’m convinced.

Step 2: Try the first bounds you can think of

We know that −1≤sin⁡(x)≤1-1 leq sin(x)leq 1−1≤sin(x)≤1, so maybe we can copy our strategy from the xsin⁡(1/x)xsin(1/x)xsin(1/x) example. If we do that, then we get

−1∣x∣≤sin⁡(x)x≤1∣x∣.begin{aligned} -frac{1}{|x|} leq frac{sin(x)}{x} leq frac{1}{|x|}. end{aligned}−∣x∣1​≤xsin(x)​≤∣x∣1​.​

Great! Only trouble is, as x→0xto 0x→0, our lower bound goes to −∞-infty−∞ and our upper bound goes to ∞,infty,∞, so we now only know that if the limit exists,

−∞≤lim⁡x→0sin⁡(x)x≤∞,begin{aligned} -inftyleq lim_{xto 0}frac{sin(x)}{x}leq infty, end{aligned}−∞≤x→0lim​xsin(x)​≤∞,​

which just means that lim⁡x→0sin⁡(x)xlim_{xto 0}frac{sin(x)}{x}limx→0​xsin(x)​ is “some number”. Not exactly helpful.

Note: You can use these bounds to prove that lim⁡x→−∞sin⁡(x)x=lim⁡x→−∞sin⁡(x)x=0lim_{xto -infty} frac{sin(x)}{x} = lim_{xto -infty}frac{sin(x)}{x} = 0limx→−∞​xsin(x)​=limx→−∞​xsin(x)​=0. You have to modify the squeeze theorem a bit though to make sense of limits at infinity.

Step 3: Get clever.

As suggested by Catherine, we’re breaking out some triangles and circles. Consider the normal setup on the unit circle, only this time we’re adding a second, bigger triangle (seen in green):

We’re going to compare the areas of these three shapes.

Area of the small triangle

Using the formula Area of Triangle=12(base)⋅(height)text{Area of Triangle} = frac 12 text{(base)}cdot text{(height)}Area of Triangle=21​(base)⋅(height), we get that the area of the small triangle is

Area⁡(A)=12cos⁡θsin⁡θ.begin{aligned} operatorname{Area}(A) = frac12 costhetasintheta. end{aligned}Area(A)=21​cosθsinθ.​

Area of the wedge

Recall that the fraction of the area taken up by a circle wedge is θ/2πtheta/2piθ/2π:

This means that

Area⁡(B)=θ2.begin{aligned} operatorname{Area}(B) = frac{theta}{2}. end{aligned}Area(B)=2θ​.​

Area of the large triangle

From the diagram, it’s

Area⁡(C)=12⋅tan⁡θ.begin{aligned} operatorname{Area}(C) = frac 12 cdot tantheta. end{aligned}Area(C)=21​⋅tanθ.​

Comparing the areas

Based on the construction of these three shapes in the original picture, we see that AAA sits inside BBB and BBB sits inside CCC. Thus

Area⁡(A)≤Area⁡(B)≤Area⁡(C).begin{aligned} operatorname{Area}(A) leq operatorname{Area}(B) leq operatorname{Area}(C). end{aligned}Area(A)≤Area(B)≤Area(C).​

Doing some algebra

Substituting the formulas we found in for the areas in inequality (15), we get

12cos⁡θsin⁡θ≤θ2≤12tan⁡θ.begin{aligned} frac12 costhetasintheta leq frac{theta}{2} leq frac12 tantheta. end{aligned}21​cosθsinθ≤2θ​≤21​tanθ.​

Multiply by 222 to get

cos⁡θsin⁡θ≤θ≤tan⁡θ=sin⁡θcos⁡θ.begin{aligned} costhetasintheta leq theta leq tantheta = frac{sintheta}{costheta}. end{aligned}cosθsinθ≤θ≤tanθ=cosθsinθ​.​

Now we make some restrictions. We’re eventually going to be applying the squeeze theorem at θ=0theta = 0θ=0, so we may as well restrict our possible values of θthetaθ. Let’s say that −π/2<θ<π/2-pi/2 < theta < pi/2−π/2<θ<π/2; if you look at the statement of the squeeze theorem, we have chosen ϵ=π/2epsilon = pi/2ϵ=π/2. Now we’re working in the interval (−π2,π2)left(-fracpi2, fracpi2right)(−2π​,2π​).

Note: We actually already implicitly made this restriction to θthetaθ. If θthetaθ was any larger, then the triangles wouldn’t have angles that sum to πpiπ radians.

I’d like to divide by sin⁡θsinthetasinθ, but in order to control what happens to the inequalities, I need to break into the cases sin⁡θ>0sintheta > 0 sinθ>0 and sin⁡θ<0sintheta < 0sinθ<0.

Appling the squeeze theorem

We’re now ready to apply the squeeze theorem. Set g(θ)=cos⁡θg(theta) = costhetag(θ)=cosθ and h(θ)=1cos⁡θh(theta) = frac{1}{costheta}h(θ)=cosθ1​. By what we have just shown with a lovely combination of geometry and algebra is that, whenever θ∈(−π2,π2)theta in left(-fracpi2,fracpi2right)θ∈(−2π​,2π​), we have

g(θ)≤sin⁡θθ≤h(θ).begin{aligned} g(theta) leq frac{sintheta}{theta}leq h(theta). end{aligned}g(θ)≤θsinθ​≤h(θ).​

Taking limits, we get

lim⁡θ→0g(θ)=lim⁡θ→0cos⁡θ=cos⁡(0)=1,begin{aligned} lim_{thetato 0} g(theta) = lim_{thetato 0} costheta = cos(0) = 1, end{aligned}θ→0lim​g(θ)=θ→0lim​cosθ=cos(0)=1,​ lim⁡θ→0h(θ)=lim⁡θ→01cos⁡θ=1cos⁡(0)=1,begin{aligned} lim_{thetato 0} h(theta) = lim_{thetato 0} frac{1}{costheta} = frac{1}{cos(0)} = 1, end{aligned}θ→0lim​h(θ)=θ→0lim​cosθ1​=cos(0)1​=1,​

and hence by the squeeze theorem we get

lim⁡θ→0sin⁡θθ=1.begin{aligned} lim_{thetato 0}frac{sintheta}{theta} = 1. end{aligned}θ→0lim​θsinθ​=1.​

We’re done! Here’s a last graph to illustrate that these bounds do indeed work, in case you don’t trust the algebra.