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I'm having some confusion on understanding the proof that

$$\lim_{x\to 0}\frac{\sin x}{x}=1.$$

In the proof on khanacademy and on other various sources we always end up with the inequality $$\cos(x)\le\frac{\sin(x)}{x}\le1\ $$ And then we apply the squeeze theorem.

However in the proof we always say that it is valid for any non zero x however if this is true how can we ever say that the inequality is less than or equal to each other my question is why isn't the inequality this instead

$$\cos(x)<\frac{\sin(x)}{x}<1 $$ No one seems to explain how they make the less than or equal to statement true so which one is correct?

Théophile
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The homeschooler
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    Because $< \implies \leq$. He who can do the more can do the less. – Bernard Dec 28 '20 at 16:38
  • Great question. The key point to note here is that the condition for applying the squeeze theorem requires $\leq$, not $<$. And as such, while the latter statement is true, it isn't necessary. – Rushabh Mehta Dec 28 '20 at 16:39
  • Oh so $a<b$ is the same as a$\le$b – The homeschooler Dec 28 '20 at 16:42
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    It's not the same, but, in this context, it has the same implication – J. W. Tanner Dec 28 '20 at 16:43
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    The inequality is not true for all non-zero $x$! At $x=2\pi$, for example, $\cos x = 1$ but $\frac{\sin x}x = 0$. – Théophile Dec 28 '20 at 16:44
  • Think about the geometric interpretations of the trigonometric functions. This link is pretty good https://www.maths.tcd.ie/~lebed/Remarkable.pdf – Stanislas CASTELLANA Dec 28 '20 at 16:44
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    @StanislasCASTELLANA: did you read the question at all ? –  Dec 28 '20 at 16:46
  • @don thousand how come I can say $\le$ even if it is not true – The homeschooler Dec 28 '20 at 16:51
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    It is true. For example, $4 \leq 6$ is a true statement. – D_S Dec 28 '20 at 16:54
  • "how come I can say ≤ even if it is not true " It is true. $a \le b$ means "either $a =b$ OR $a < b$. And if $a < b$ then it is certainly true that $a =b$ or $a < b$. The only way $a \le b$ can be false is if $a$ is neither equal to $b$ nor less than $b$. As $a$ is less than $b$ it is not false and is therefore true. .... Suppose someone said "Arthur lives somewhere in the United states". Then some one said "Arthur lives two miles east of the UCLA campus in Los Angeles, Ca". Does that mean we can't say "Arthur lives somewhere in the United States" if we actually know more? – fleablood Jan 01 '21 at 07:02
  • I think they didn't say $\forall x\ne 0$, but $0<|x|<\frac\pi4$. Related question. – PinkyWay Jan 01 '21 at 22:04

3 Answers3

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The answers thus far are missing an important point. It's not merely allowed to switch to the weaker, $\leq$ inequality; it is necessary to do so.

Suppose $f(x) < g(x)$ for all $x$, and that $\lim_{x\to a}f(x)$ and $\lim_{x\to a}g(x)$ both exist.

In this case it is true that $\lim_{x\to a}f(x) \leq \lim_{x\to a}g(x)$.

However, It is not necessarily true that $\lim_{x\to a}f(x) < \lim_{x\to a}g(x)$.

As an example, consider

$$f(x) = 0$$

$$g(x) = |x| \text{ (if $x\neq 0$)}$$ $$g(x) = 1 \text{ (if $x = 0$)}$$

Here, $f(x) < g(x)$ for all $x$ but $\lim_{x\to 0}f(x) = 0 = \lim_{x\to 0}g(x)$

Frequently, when taking limits on both sides of a strict inequality, we're forced to weaken the inequality from $<$ to $\leq$.

In the case of $\cos(x)<\frac{\sin(x)}{x}<1$, if we wrongly insist that strict inequality holds for the limits of the functions, we end up with $1 < \lim_{x\to 0}\frac{\sin(x)}{x} < 1$, which is clearly nonsense.

There's an exercise in Calculus by Spivak that explores some of this in more detail: chapter 5, exercise 12 (3rd ed.). I mention it here because I know @The homeschooler has access to Spivak :). Here are the relevant parts:

5-12.(a) Suppose that $f(x) \leq g(x)$ for all $x$. Prove that $\lim_{x\to a} f(x) \leq \lim_{x\to a} g(x)$, provided that these limits exist.

(c) If $f(x) < g(x)$ for all $x$, does it necessarily follow that $\lim_{x \to a} f(x) < \lim_{x \to a} g(x)$?

For part (a), you can begin by assuming $\lim_{x\to a} f(x) > \lim_{x\to a} g(x)$ and then showing how this leads to a contradiction.

Part (c) can be dealt with by giving any counter-example (the $f$ and $g$ mentioned above will work, and are from the Answer Book). Part (c) emphasizes that $\leq$ is the best we can do; we can show that $\lim_{x\to a} f(x) \not> \lim_{x\to a} g(x)$, and that is all. We cannot show strict inequality holds for the limits, even though it holds for the functions for all $x$.

Though this exercise is slightly different than the case homeschooler is asking about, the arguments are largely applicable.

Edit: I've neglected an important point myself! There is a moment when we do actively choose to weaken the inequality.

Given $\cos(x)<\frac{\sin(x)}{x}<1$ in some neighborhood around $0$, the above arguments can get us to $1\leq \lim_{x \to 0}\frac{\sin(x)}{x} \leq 1$, but only if we assume that this middle limit exists, which of course we shouldn't do because that's what we're trying to prove!

Generally speaking, if $f(x) \leq g(x) \leq h(x)$ and $\lim_{x \to a} f(x) = \lim_{x \to a} h(x)$, then $\lim_{x \to a} g(x)$ exists and is equal to the other two.

If we weaken the inequalities to read $\cos(x)\leq \frac{\sin(x)}{x} \leq 1$, we can use the above theorem to get $\lim_{x \to 0}\frac{\sin(x)}{x} = 1$, all while avoiding any assumptions about existence.

Proving this general "limits+squeeze theorem" theorem is Spivak's next exercise, 5-13.

Suppose $f(x) \leq g(x) \leq h(x)$ and $\lim_{x \to a} f(x) = L = \lim_{x \to a} h(x)$.

For any $\varepsilon > 0$ we can find a $\delta > 0$ such that for all $x$, if $0 < |x - a| < \delta$, then $L - \varepsilon < f(x) < L + \varepsilon$, and also $L - \varepsilon < h(x) < L + \varepsilon$.

Therefore we have

$$L - \varepsilon < f(x) \leq g(x) \leq h(x) < L + \varepsilon$$

which shows $|g(x) - L| < \varepsilon$.

Anyway, sorry for the confusion and I'm guessing I've managed somehow to provide simultaneously too much and not enough in this answer.

Edit 2: reading it over again, I don't think we actually need to choose to weaken the inequalities for $\cos(x)<\frac{\sin(x)}{x}<1$. The above "limits+squeeze theorem" proof works even if we replace all the $\leq$'s with strict inequalities.

I'm relieved. Choosing to weaken it is too clever for me.

In summary,

(1) You can take limits of both sides of an inequality if you know the limits exist, but doing so turns strict $<$ into $\leq$. This is because...

(2) Two functions that are unequal everywhere can still have equal limits.

(3) If you have some function sandwiched between two other functions and those functions have the same limit, then so does the middle one.

(4) The requirement "...if $a \leq b$..." is certainly met by $a <b$.

Ben
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    +1 for the well-written and needed explanation @Ben! – ultralegend5385 Jan 01 '21 at 07:24
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    Cheers ultralegend5385, and happy new year! – Ben Jan 01 '21 at 07:40
  • Thankyou @ben this was definitely the clearest explanation – The homeschooler Jan 01 '21 at 10:17
  • Very happy new year @Ben! – ultralegend5385 Jan 01 '21 at 11:37
  • You're welcome @The homeschooler. Sometimes you'll see an inequality with $<$. You'll then see an operation in which the writer "takes the limit of both sides" of the inequality, and $<$ suddenly turns into $\leq$. – Ben Jan 01 '21 at 13:49
  • To "take the limit of both sides" of an inequality, what we're really doing is noting that the function on the LHS of the inequality is $<$(or $>$,$\leq$, or$\geq$) than the function on the RHS in a neighborhood containing the point where the limit is to be taken (but possibly excluding the point itself), and also assuming the limits both exist. These details are often suppressed and so it seems baffling when $<$ seems to turn into $\leq$ "for no reason." – Ben Jan 01 '21 at 14:09
  • To avoid having to explain these complications, I suspect many calculus textbooks (and MSE posters ;P) instead go with the easier seeming "the squeeze theorem needs $\leq$ and since we have $<$, we have $\leq$" explanation. – Ben Jan 01 '21 at 14:16
  • Yes indeed they do which confused me a lot for a while – The homeschooler Jan 01 '21 at 14:30
  • Best answer by far. – K.defaoite Jan 01 '21 at 20:25
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    I really shouldn't have butted in. I haven't yet gotten to Spivak's chapter where he formally defines the trig functions. Ah well, in trying to punch above my weight here, I think I clarified my own understanding. – Ben Jan 01 '21 at 20:33
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$a<b$ is obviously not the same as $a \leq b$. However, if we know that $a<b$, then it is of course true that $a \leq b$.

D_S
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$$a\le b$$ means $$a<b\text{ or }a=b.$$

It suffices that one of the conditions be true for the whole expression to be true.

You seem to interpret that $a\ne b$ invalidates it, but this is wrong.