Before we use the formal definition, let's try some numerical tolerances. What if the \(y\) tolerance is \(0.5\text{,}\) or in other words \(\varepsilon =0.5\text{?}\) How close to \(4\) does \(x\) have to be so that \(y\) is within \(0.5\) units of \(2\text{?}\) That is, \(1.5 \lt y \lt 2.5\text{?}\) In this case, we can proceed as follows: \begin{align*} 1.5 \amp\lt y \lt 2.5\\ 1.5 \amp\lt \sqrt{x} \lt 2.5 \amp\amp \text{ (Let } y=\sqrt{x} \text{)}\\ 1.5^2 \amp\lt x \lt 2.5^2\amp\amp \text{(Square the inequality)}\\ 2.25 \amp\lt x \lt 6.25\\ 2.25-4 \amp\lt x-4 \lt 6.25-4\amp\amp \text{(Subtract }4\text{ from both sides)}\\ -1.75 \amp\lt x-4 \lt 2.25 \end{align*}

So, what is the desired \(x\) tolerance? Remember, we want to find a \(\delta\) so that \(\abs{x-4}\) is smaller than \(\delta\text{.}\) Since \(1.75\lt2.25\text{,}\) then if we require \(\abs{x-4}\lt 1.75\text{,}\) then we have \begin{align*} \amp\abs{x-4}\lt1.75\\ \implies\amp-1.75\lt x-4\lt1.75\lt2.25 \end{align*} Therefore we can have \(\delta \leq 1.75\text{.}\) See Figure 1.2.3.

Given the \(y\) tolerance \(\varepsilon =0.5\text{,}\) we have found an \(x\) tolerance, \(\delta \leq 1.75\text{,}\) such that whenever \(x\) is within \(\delta\) units of \(4\text{,}\) then \(y\) is within \(\varepsilon\) units of \(2\text{.}\) That's what we were trying to find.

Let's try another value of \(\varepsilon\text{.}\)

What if the \(y\) tolerance is \(0.01\text{,}\) i.e. \(\varepsilon =0.01\text{?}\) How close to \(4\) does \(x\) have to be in order for \(y\) to be within \(0.01\) units of \(2\text{?}\) (In other words for \(1.99 \lt y \lt 2.01\text{?}\)) Again, we just square these values to get \(1.99^2 \lt x \lt 2.01^2\text{,}\) or \begin{align*} 3.9601 \amp\lt x \lt 4.0401\\ -0.0399\amp\lt x-4 \lt 0.0401 \end{align*}

What is the desired \(x\) tolerance? In this case we must have \(\delta \leq 0.0399\text{,}\) which is the minimum distance from \(4\) of the two bounds given above.

What we have so far: if \(\varepsilon =0.5\text{,}\) then \(\delta \leq 1.75\) leads to \(f(x)\) being less than \(\varepsilon\) from \(f(4)\) and if \(\varepsilon = 0.01\text{,}\) then \(\delta \leq 0.0399\) being less than \(\varepsilon\) from \(f(4)\text{.}\) A pattern is not easy to see, so we switch to general \(\varepsilon\) try to determine an adequate \(\delta\) symbolically. We start by assuming \(y=\sqrt{x}\) is within \(\varepsilon\) units of \(2\text{:}\) \begin{align*} \amp\abs{y - 2} \lt \varepsilon\\ -\varepsilon \amp\lt y - 2 \lt \varepsilon\\ -\varepsilon \amp\lt \sqrt{x} - 2 \lt \varepsilon \amp\amp (y=\sqrt{x})\\ 2 - \varepsilon \amp\lt \sqrt{x} \lt 2+ \varepsilon \amp\amp \text{(Add 2)}\\ (2 - \varepsilon)^2 \amp\lt x \lt (2+ \varepsilon) ^2 \amp\amp \text{(Square all)}\\ 4 - 4\varepsilon + \varepsilon^2 \amp\lt x \lt 4 + 4\varepsilon + \varepsilon^2 \amp\amp \text{(Expand)}\\ -4\varepsilon + \varepsilon^2 \amp\lt x-4 \lt 4\varepsilon + \varepsilon^2\amp\amp \text{(Subtract 4)} \end{align*}

We choose the smaller of these two distances from the number \(4\text{:}\) \begin{equation*} \delta \leq \min\{4\varepsilon - \varepsilon^2, 4\varepsilon + \varepsilon^2\}\text{.} \end{equation*}

Since \(\varepsilon > 0\text{,}\) we have \(4\varepsilon - \varepsilon^2 \lt 4\varepsilon + \varepsilon^2\text{,}\) the minimum is \(\delta \leq 4\varepsilon - \varepsilon^2\text{.}\) That's the formula: given an \(\varepsilon\text{,}\) set \(\delta \leq 4\varepsilon-\varepsilon^2\text{.}\)

We can check this for our previous values. If \(\varepsilon=0.5\text{,}\) the formula gives \(\delta \leq 4(0.5) - (0.5)^2 = 1.75\) and when \(\varepsilon=0.01\text{,}\) the formula gives \(\delta \leq 4(0.01) - (0.01)^2 = 0.399\text{.}\)

So given any \(\varepsilon >0\text{,}\) set \(\delta \leq 4\varepsilon - \varepsilon^2\text{.}\) Then if \(\abs{x-4}\lt \delta\) (and \(x\neq 4\)), then \(\abs{f(x) - 2} \lt \varepsilon\text{,}\) satisfying the definition of the limit. We have shown formally (and finally!) that \(\lim\limits_{x\to 4} \sqrt{x} = 2\text{.}\)

Let's do this example symbolically from the start. Let \(\varepsilon > 0\) be given; we want \(\abs{y-4} \lt \varepsilon\text{,}\) i.e., \(\abs{x^2-4} \lt \varepsilon\text{.}\) How do we find \(\delta\) such that when \(\abs{x-2} \lt \delta\text{,}\) we are guaranteed that \(\abs{x^2-4}\lt \varepsilon\text{?}\)

This is a bit trickier than the previous example, but let's start by noticing that \(\abs{x^2-4} = \abs{x-2}\cdot\abs{x+2}\text{.}\) Consider: \begin{equation*} \abs{x^2-4} \lt \varepsilon \implies \abs{x-2}\cdot\abs{x+2} \lt \varepsilon \implies \abs{x-2} \lt \frac{\varepsilon}{\abs{x+2}}\text{.} \end{equation*}

Could we not set \(\delta = \frac{\varepsilon}{\abs{x+2}}\text{?}\)

We are close to an answer, but the catch is that \(\delta\) must be a constant value (so it can't depend on \(x\)). There is a way to work around this, but we do have to make an assumption. Remember that \(\varepsilon\) is supposed to be a small number, which implies that \(\delta\) will also be a small value. In particular, we can (probably) assume that \(\delta \lt 1\text{.}\) If this is true, then \(\abs{x-2} \lt \delta\) would imply that \(\abs{x-2} \lt 1\text{,}\) giving \(1 \lt x \lt 3\text{.}\)

Now, back to the fraction \(\frac{\varepsilon}{\abs{x+2}}\text{.}\) If \(1\lt x\lt 3\text{,}\) then \(3\lt x+2\lt 5\) (add \(2\) to all terms in the inequality). Taking reciprocals, we have \begin{equation*} \frac15 \lt \frac1{\abs{x+2}} \lt \frac 13\text{,} \end{equation*} which implies \begin{equation*} \frac15 \lt \frac1{\abs{x+2}}\text{,} \end{equation*} which implies \begin{equation} \frac\varepsilon5\lt \frac{\varepsilon}{\abs{x+2}}\text{.}\label{eq_limit1}\tag{1.2.1} \end{equation}

This suggests that we set \(\delta \leq \frac{\varepsilon}{5}\text{.}\) To see why, let consider what follows when we assume \(\abs{x-2}\lt \delta\text{:}\) \begin{align*} \abs{x-2}\amp\lt\delta\\ \abs{x-2}\amp\lt\frac{\varepsilon}{5}\amp\amp\text{(Our choice of }\delta\text{)}\\ \abs{x-2}\cdot\abs{x+2}\amp\lt\abs{x+2}\cdot\frac{\varepsilon}{5}\amp\amp\text{(Multiply by }\abs{x+2}\text{)}\\ \abs{x^2-4}\amp\lt\abs{x+2}\cdot\frac{\varepsilon}{5}\amp\amp\text{(Simplify left side)}\\ \abs{x^2-4}\amp\lt\abs{x+2}\cdot\frac{\varepsilon}{\abs{x+2}}\amp\amp\text{(}\knowl{./knowl/eq_limit1.html}{\text{Inequality (1.2.1)}}, \delta \lt 1\text{)}\\ \abs{x^2-4}\amp\lt\varepsilon \end{align*}

We have arrived at \(\abs{x^2-4}\lt \varepsilon\) as desired. Note again, in order to make this happen we needed \(\delta\) to first be less than \(1\text{.}\) That is a safe assumption; we want \(\varepsilon\) to be arbitrarily small, forcing \(\delta\) to also be small.

We have also picked \(\delta\) to be smaller than “necessary.” We could get by with a slightly larger \(\delta\text{,}\) as shown in Figure 1.2.5. The outer lines show the boundaries defined by our choice of \(\varepsilon\text{.}\) The inner lines show the boundaries defined by setting \(\delta = \varepsilon/5\text{.}\) Note how these dotted lines are within the dashed lines. That is perfectly fine; by choosing \(x\) within the dotted lines we are guaranteed that \(f(x)\) will be within \(\varepsilon\) of \(4\text{.}\)

In summary, given \(\varepsilon > 0\text{,}\) set \(\delta=\varepsilon/5\text{.}\) Then \(\abs{x-2} \lt \delta\) implies \(\abs{x^2 - 4}\lt\varepsilon\) (i.e. \(\abs{y-4}\lt\varepsilon\)) as desired. This shows that \(\lim\limits_{x\to 2} x^2 = 4\text{.}\) Figure 1.2.5 gives a visualization of this; by restricting \(x\) to values within \(\delta = \varepsilon/5\) of \(2\text{,}\) we see that \(f(x)\) is within \(\varepsilon\) of \(4\text{.}\)

We start our scratch-work by considering \(\abs{f(x) - (-1)} \lt \varepsilon\text{:}\) \begin{align} \abs{f(x)-(-1)} \amp \lt \varepsilon\notag\\ \abs{x^3-2x + 1}\amp \lt \varepsilon \amp\amp \text{(Now factor)}\notag\\ \abs{(x-1)(x^2+x-1)}\amp \lt \varepsilon\notag\\ \abs{x-1}\amp\lt\frac{\varepsilon}{\abs{x^2+x-1}\text{.}}\label{eq_lim4}\tag{1.2.2} \end{align}

We are at the phase of saying that \(\abs{x-1}\lt\) something, where something\(=\varepsilon/\abs{x^2+x-1}\text{.}\) We want to turn that something into \(\delta\text{.}\)

Since \(x\) is approaching \(1\text{,}\) we are safe to assume that \(x\) is between \(0\) and \(2\text{.}\) So \begin{align*} 0\amp \lt x\lt 2\\ 0\amp \lt x^2\lt 4.\amp\amp \text{(squared each term)}\\ \end{align*} Since \(0\lt x\lt 2\text{,}\) we can add \(0\text{,}\) \(x\) and \(2\text{,}\) respectively, to each part of the inequality and maintain the inequality. \begin{align*} 0\amp \lt x^2+x\lt 6\\ -1\amp \lt x^2+x-1\lt 5\text{.}\amp\amp \text{(subtracted 1 from each part)} \end{align*}

In Inequality (1.2.2), we wanted \(\abs{x-1}\lt \varepsilon/\abs{x^2+x-1}\text{.}\) The above shows that given any \(x\) in \([0,2]\text{,}\) we know that \begin{align} x^2+x-1 \amp \lt 5 \amp\amp \text{which implies that}\notag\\ \frac15 \amp \lt \frac{1}{x^2+x-1} \amp\amp \text{which implies that}\notag\\ \frac{\varepsilon}5 \amp \lt \frac{\varepsilon}{x^2+x-1}\text{.}\label{eq_lim5}\tag{1.2.3} \end{align}

So we set \(\delta \leq \varepsilon/5\text{.}\) This ends our scratch-work, and we begin the formal proof (which also helps us understand why this was a good choice of \(\delta\)).

Given \(\varepsilon\text{,}\) let \(\delta \leq \varepsilon/5\text{.}\) We want to show that when \(\abs{x-1}\lt \delta\text{,}\) then \(\abs{(x^3-2x)-(-1)}\lt \varepsilon\text{.}\) We start with \(\abs{x-1}\lt \delta\text{:}\) \begin{align*} \abs{x-1} \amp \lt \delta\\ \abs{x-1} \amp \lt \frac{\varepsilon}{5}\\ \abs{x-1} \amp \lt \frac{\varepsilon}{\abs{x^2+x-1}} \amp\amp \text{(}\knowl{./knowl/mdn-2.html}{\text{Inequality (1.2.3)}}, x\text{ near 1)}\\ \abs{x-1}\cdot \abs{x^2+x-1} \amp \lt \varepsilon\\ \abs{x^3-2x+1} \amp \lt \varepsilon\\ \abs{(x^3-2x)-(-1)} \amp \lt \varepsilon, \end{align*} which is what we wanted to show. Thus \(\lim\limits_{x\to 1}x^3-2x = -1\text{.}\)

Symbolically, we want to take the inequality \(\abs{e^x - 1} \lt \varepsilon\) and unravel it to the form \(\abs{x-0} \lt \delta\text{.}\) Here is our scratch-work: \begin{align*} \amp\abs{e^x - 1}\lt \varepsilon\\ -\varepsilon \amp\lt e^x - 1 \lt \varepsilon\amp\amp \text{(Definition of absolute value)}\\ 1-\varepsilon \amp\lt e^x \lt 1+\varepsilon \amp\amp \text{(Add 1)}\\ \ln(1-\varepsilon) \amp\lt x \lt \ln(1+\varepsilon) \amp\amp \text{(Take natural logs)} \end{align*}

Making the safe assumption that \(\varepsilon\lt 1\) ensures the last inequality is valid (i.e., so that \(\ln(1-\varepsilon)\) is defined). We can then set \(\delta\) to be the minimum of \(\abs{\ln(1-\varepsilon)}\) and \(\ln(1+\varepsilon)\text{;}\) i.e., \begin{equation*} \delta = \min\{\abs{\ln(1-\varepsilon)}, \ln(1+\varepsilon)\} = \ln(1+\varepsilon). \end{equation*}

Recall \(\ln(1-\varepsilon)\) is negative because \(1-\varepsilon\lt1\text{.}\) So \(\abs{\ln(1-\varepsilon)}=-\ln(1-\varepsilon)=\ln\left(\frac{1}{1-\varepsilon}\right)\text{.}\) So to determine which is smaller between \(\abs{\ln(1-\varepsilon)}\) and \(\ln(1+\varepsilon)\) amounts to determining which is smaller between \(\frac{1}{1-\varepsilon}\) and \(1+\varepsilon\text{.}\) But \((1+\varepsilon)/\left(\frac{1}{1-\varepsilon}\right)=(1+\varepsilon)(1-\varepsilon)=1-\varepsilon^2\lt1\text{,}\) so \((1+\varepsilon)\lt\frac{1}{1-\varepsilon}\text{.}\) And therefore \(\ln(1+\varepsilon)\lt\abs{\ln(1-\varepsilon)}\text{.}\)

Now, we work through the actual the proof: \begin{align*} \amp\abs{x - 0}\lt \delta\\ -\delta \amp \lt x \lt \delta \amp\amp \text{(Definition of absolute value)}\\ -\ln(1+\varepsilon) \amp \lt x \lt \ln(1+\varepsilon)\\ \ln(1-\varepsilon) \amp \lt x \lt \ln(1+\varepsilon)\text{.}\amp \amp \text{(since \(\ln(1-\varepsilon) \lt -\ln(1+\varepsilon)\)) }\\ \end{align*} The above line is true by our choice of \(\delta\) and by the fact that since \(\abs{\ln(1-\varepsilon)}>\ln(1+\varepsilon)\) and \(\ln(1-\varepsilon)\lt 0\text{,}\) we know \(\ln(1-\varepsilon) \lt -\ln(1+\varepsilon )\text{.}\) \begin{align*} 1-\varepsilon \amp \lt e^x \lt 1+\varepsilon \amp \amp \text{(Exponentiate)}\\ -\varepsilon \amp \lt e^x - 1 \lt \varepsilon \amp\amp \text{(Subtract 1)} \end{align*}

In summary, given \(\varepsilon > 0\text{,}\) let \(\delta = \ln(1+\varepsilon)\text{.}\) Then \(\abs{x - 0} \lt \delta\) implies \(\abs{e^x - 1}\lt \varepsilon\) as desired. We have shown that \(\lim\limits_{x\to 0} e^x = 1\text{.}\)