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I have been examining many different examples and I found no objective justification to the chosen bounds in none of them, as if the choice was an intuitive process. Is it really just that? For the lower bound, do we "begin with" a 0 and start looking for a bound that is the furthest possible from 0 and that still satisfies the inequality? What about the upper bound? It just seems to me that it is very easy to choose the wrong bounds given the commonly counterintuitive nature of limits for beginners. Is there a way to be sure that it is the right/not right one and not fall onto that? (In case it makes a difference, the examples I have examined were applied on sequences)

Any help is welcome, thank you.

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    $\begingroup$ My two cents: experience and intuition. $\endgroup$ – Lucas Henrique Jan 26 at 0:38
  • $\begingroup$ Both the upper and lower bound should converge to the same limit... If they don't, then I know I have picked the wrong bounds. If you're working on a sequence, a common trick is to replace the variable terms in each sequence with a constant to get a sum that is larger/smaller than said limit. $\endgroup$ – Aniruddha Deb Jan 26 at 2:27
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The Squeeze Theorem

So just for context, here's the definition of the Squeeze Theorem my answer works with. There are more general versions of the theorem, which correspond to "convergence" in metric or topological spaces, but we'll just use a version for the real numbers to keep things simple.

Squeeze Theorem: Let $f : X\to\mathbb{R}$, and let $f^{-} : X\to\mathbb{R}$ and $f^{+} : X\to\mathbb{R}$, be functions defined on some set $X$, and let $a \in \overline{X}$ be a limit point of $X$ (but not necessarily in $X$ itself). Then if $$ \forall x\in X : f^{-}(x) \le f(x) \le f^{+}(x) $$ And if additionally, $$ \lim_{x\to a}f^{-}(x) = \lim_{x\to a}f^{+}(x) = L $$ Then $$ \lim_{x\to a}f(x) = L $$

The theorem can apply to functions $\mathbb{R}\to\mathbb{R}$, if you take $x \in X \subseteq \mathbb{R}$ to be a point on some interval of the real numbers, and $f$ to be a real-valued function.

Or, it can also apply equally to number sequences, by taking $X = \mathbb{N}$ to be a the set of natural numbers, taking $x = n \in X$ to be some natural number, and taking $f(n) = s_n$ to be the function which maps each sequence index $n = \{1, 2, 3, ...\}$ to the $n$th sequence element $s_n = \{s_1, s_2, s_3, ...\}$. In this case, the point $a=\infty$ is not in $\mathbb{N}$, but is still one of the limit points of $\mathbb{N}$.

Now, to get to your question: it does not matter how fast the squeeze happens, only that there is a pinch. At the beginning the lower and upper bounding sequences can be as "lower" and "upper" as you like, as long as they stay below/above $f(x)$ and both converge to $L$. What matters is that eventually, $f^{-}(x)$ and $f^{+}(x)$ will "squeeze" $f(x)$ to the same point, because $f$ is between $f^{-}$ and $f^{+}$ at every single step of the way, but $f^{-}$ and $f^{+}$ are converging to the same limit $L$, so in essence eventually there is "nowhere else to go" for $f$ except for also to the same limit $L$.


An example with functions

In cases where $|a| < \infty$, this squeezing action can be visualized readily. Here is an example with $a = 0$:

Squeeze Theorem - Example

There is also an interactive graph you can use to explore this example further.

In this case, $f(x) = 3x^3\sin\left(1/x^2\right)$ is the black curve, which is undefined at $x=0$. We can take $X = (-1, 1)$ or some other interval around $x=0$ (since we're interested in what happens as $x\to 0$, so it really doesn't matter what these functions do far away from $x=0$).

  1. If you take the lower and upper limiting functions to be $f^{-}(x) = -3|x|^3$, the solid blue curve, and $f^{+}(x) = 3|x|^3$, the solid red curve, then the conditions of the squeeze theorem are satisfied: \begin{array}{cc} \forall{x}\in X : -3|x|^3 \le 3x^3\sin\left(1/x^2\right) \le 3|x|^3 & \checkmark \\ \lim_{x\to 0}\, -3|x|^3 = \lim_{x\to 0}\, 3|x|^3 = -3|0|^3 = 3|0|^3 = 0 & \checkmark \end{array}

    Therefore, by the Squeeze Theorem, the black curve also converges to $L = 0$: $$ \lim_{x\to 0}\; 3x^3\sin\left(1/x^2\right) = 0 $$ However, you may notice something about the solid red and black curves - they actually come close enough to the black curve to just barely touch it at several points. In a sense, they are the "tightest" possible lower and upper bounds you can place on $f$.

    Many illustrations of the Squeeze Theorem use "tight" bounds of this sort, perhaps because visually it emphasizes the "squeezing" action for which the theorem is named. But it is not necessary for the bounds to be exactly tight, for the theorem to work!

  2. We could also take the lower and upper limiting functions to be $g^{-}(x) = -3x^2$, the dashed blue curve, and $g^{+}(x) = 3x^2$, the dashed red curve. Notice that the bound is no longer as tight, but the conditions of the squeeze theorem are satisfied all the same! \begin{array}{cc} \forall{x}\in X : -3x^2 \le 3x^3\sin\left(1/x^2\right) \le 3x^2 & \checkmark \\ \lim_{x\to 0}\, -3x^2 = \lim_{x\to 0}\, 3x^2 = -3(0)^2 = 3(0)^2 = 0 & \checkmark \end{array} These functions would work just as well for establishing the limit of $f(x)$ as $x\to 0$ as the first pair.


An example with sequences

Here's another example, this time with a discrete sequence. Consider the infinite sequence: $$ a_n = \frac{(-1)^n}{n!} = \left(-1, \frac{1}{2}, -\frac{1}{6}, \frac{1}{24}, -\frac{1}{120}, \cdots \right) $$ We can use the Squeeze Theorem to show that $a_n \to 0$ as $n \to \infty$, by finding two sequences $a^{\;-}_{n}$ and $a^{\;+}_n$, where both (1) $a^{\;-}_n \le a_n \le a^{\;+}_n$ for all $n$, and (2) both $a^{\;-}_n \to 0$ and $a^{\;+}_n \to 0$ as $n \to 0$. But just like with the continuous example, the bounding sequences don't have to be "tight" limits - they just have to bound, and to converge to the same limit. Therefore you could use any of the following as the Squeeze Theorem lower/upper bounds: \begin{array}{cll} \text{(1)} & a^{\;-}_n = -1/n! \;&\; a^{\;+}_n = 1/n! \\ \text{(2)} & a^{\;-}_n = -1/2^{n-1} \;&\; a^{\;+}_n = 1/2^{n-1} \\ \text{(3)} & a^{\;-}_n = -1/n \;&\; a^{\;+}_n = 1/n \\ \end{array}

Only the sequences in item $\text{(1)}$ bound the sequence $a_n$ "tightly". Nevertheless, any of these sequences would work to establish the limit of $a_n$ through the Squeeze Theorem, because they all bound $a_n$, and they all converge to $0$.

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  • $\begingroup$ I found this to be very helpful, thank you! $\endgroup$ – houda el fezzak Jan 26 at 10:42
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But, we have $f(x)\le g(x)\le h(x)$ with $f$ and $h$ having the same limit. So it really is a "tight squeeze".

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