## How To Construct a Delta-Epsilon Proof

The proof, using delta and epsilon, that a function has a limit will mirror the definition of the limit. Therefore, we first recall the definition:

 $\lim\limits_{x\to c} f(x)=L$   means that for every   $\epsilon>0$,   there exists a   $\delta>0$,   such that for every $x$, the expression   $0< |x-c|<\delta$   implies   $|f(x)-L|<\epsilon$.

Each phrase of the definition contributes to some aspect of the proof. Specifically:

• The phrase "for every   $\epsilon >0$ "  implies that we have no control over epsilon, and that our proof must work for every epsilon.
• The phrase "there exists a   $\delta >0$ "   implies that our proof will have to give the value of delta, so that the existence of that number is confirmed. Typically, the value of delta will depend on the value of epsilon.
• The phrase "such that for every $x$" implies that we cannot restrict the values of $x$ any further than the next restriction provides.
• The phrase "the expression   $0< |x-c| < \delta$ "   is the starting point for a series of implications (algebra steps) which will conclude with the final statement. The expression   $|x-c| < \delta$   means that the values of $x$ will be close to $c$, specifically not more than (nor even equal to) delta units away. The expression   $0 < |x-c|$   implies that $x$ is not equal to $c$ itself.
• The phrase "implies   $|f(x)-L| < \epsilon$ "   is the conclusion of the series of implications. Once this statement is reached, the proof will be complete.

Upon examination of these steps, we see that the key to the proof is the identification of the value of delta. To find that delta, we typically begin with the final statement   $|f(x)-L| < \epsilon$,   and work backwards until we reach the form   $|x-c| < \delta$.   So let's consider some examples. Linear examples are the easiest. Non-linear examples exhibit a few other quirks, and we will demonstrate them below also.

### Example using a Linear Function

Prove, using delta and epsilon, that   $\lim\limits_{x\to 4} (5x-7)=13$.

We will place our work in a table, so we can provide a running commentary of our thoughts as we work.

 $|f(x)-L| < \epsilon$ Before we can begin the proof, we must first determine a value for delta. To find that delta, we begin with the final statement and work backwards. $|(5x-7)-13| < \epsilon$ We substitute our known values of $f(x)$ and $L$. $|5x-20| < \epsilon$ Our short-term goal is to obtain the form   $|x-c| < \delta$.   So we begin by simplifying inside the absolute value. $|5(x-4)|<\epsilon$$|5||x-4|<\epsilon |x-4|<\dfrac{\epsilon}{5} In these three steps, we divided both sides of the inequality by 5. Most often, these steps will be combined into a single step. However, when the slope of the linear function is negative, you may want to do the steps separately so as to avoid incorrectly handling the negative sign. |x-c|<\delta$$\delta=\dfrac{\epsilon}{5}$ We now recall that we were evaluating a limit as $x$ approaches 4, so we now have the form   $|x-c| < \delta$.   Therefore, since $c$ must be equal to 4, then delta must be equal to epsilon divided by 5 (or any smaller positive value).

Now we are ready to write the proof. Once again, we will provide our running commentary.

Proof.

 Suppose   $\epsilon >0$   has been provided. This is always the first line of a delta-epsilon proof, and indicates that our argument will work for every epsilon. Define   $\delta=\dfrac{\epsilon}{5}$. Since the definition of the limit claims that a delta exists, we must exhibit the value of delta. We use the value for delta that we found in our preliminary work above. Since   $\epsilon >0$,   then we also have   $\delta >0$. The definition does place a restriction on what values are appropriate for delta (delta must be positive), and here we note that we have chosen a value of delta that conforms to the restriction. Now, for every $x$, the expression   $0< |x-c| < \delta$   implies This is the next part of the wording from the definition of the limit. $|x-4| < \dfrac{\epsilon}{5}$ We replace the values of $c$ and delta by the specific values for this problem. From here on, we will be basically following the steps from our preliminary work, but in reverse order. $|5x-20| < \epsilon$ We multiplied both sides by 5. If the slope of the original function was negative, we may want to do this using more steps, so as to introduce the negative sign correctly. $|(5x-7)-13| < \epsilon$ Now we break the expression into the two parts we need to exhibit, the original function and the limit value. Therefore,   $\lim\limits_{x\to 4} (5x-7)=13$. Since we began with   $c = 4$, and we obtained the above limit statement, we have met all of the requirements of the definition of the limit, and obtained our final result. Q.E.D. This is an abbreviation for the Latin expression "quod erat demonstrandum", which means "which was to be demonstrated". Some authors will include it to denote the end of the proof.

### Example using a Non-Linear Function

Prove, using delta and epsilon, that   $\lim\limits_{x\to 5} (3x^2-1)=74$.

 $|f(x)-L| < \epsilon$ Before we can begin the proof, we must first determine a value for delta. To find that delta, we begin with the final statement and work backwards. $|(3x^2-1)-74| < \epsilon$ We substitute our known values of $f(x)$ and $L$. $|3x^2-75| < \epsilon$ Our short-term goal is to obtain the form   $|x-c| < \delta$.   However, with non-linear functions, it is easier to work toward solving for $x$ by itself, then introduce the value of $c$. So we begin by simplifying inside the absolute value. $-\epsilon < 3x^2-75 < \epsilon$ With non-linear functions, the absolute values will have to be removed, since the allowable delta-distances will be different on the two sides of the value   $x=c$. $75-\epsilon < 3x^2 < 75+\epsilon$ $25-\dfrac{\epsilon}{3} < x^2 < 25+\dfrac{\epsilon}{3}$ $\sqrt{25-\dfrac{\epsilon}{3}} < x < \sqrt{25+\dfrac{\epsilon}{3}}$ In these three steps, we solve for the variable $x$, by first adding 75 to each expression, then dividing each expression by 3, and finally taking the square root of each expression. The square root function is increasing on all real numbers, so the inequality does not change direction. If you are using a decreasing function, the inequality signs will switch direction. Notice that the two ends of the inequality are no longer opposites of one another, which means that absolute values could not be used to write these as a single inequality. Also, the left hand expression can be undefined for some values of epsilon, so we must be careful in defining epsilon. $-5+\sqrt{25-\dfrac{\epsilon}{3}} < x-5 < -5+\sqrt{25+\dfrac{\epsilon}{3}}$ Since our short-term goal was to obtain the form   $|x-c|<\delta$,   and the value of $c$ is 5, we subtract 5 from each expression. $|x-c|<\delta$$-\delta < x-c < \delta Since the two ends of the expression above are not opposites of one another, we cannot put the expression back into the form |x-c|<\delta. Therefore, we shall expand this absolute value expression instead. \delta=\min\left\{5-\sqrt{25-\dfrac{\epsilon}{3}},-5+\sqrt{25+\dfrac{\epsilon}{3}}\right\} There are two candidates for delta, and we need delta to be less than or equal to both of them. Therefore, we will require that delta be equal to the minimum of the two quantities. Notice that since the left-end expression was equivalent to negative delta, we used its opposite in our definition of delta. However, since the first candidate is undefined for \epsilon > 75, we will need to handle the "large epsilon" situation by introducing a second, smaller epsilon in the proof. Now, we are ready to write the proof. Proof.  Suppose \epsilon >0 has been provided. This is always the first line of a delta-epsilon proof, and indicates that our argument will work for every epsilon. Let \epsilon_2=\min\{\epsilon, 72\}. To avoid an undefined delta, we introduce a slightly smaller epsilon when needed. In this example, the value of 72 is somewhat arbitrary, but does need to be smaller than 75. Define \delta=\min\left\{5-\sqrt{25-\dfrac{\epsilon}{3}},-5+\sqrt{25+\dfrac{\epsilon}{3}}\right\}. Since the definition of the limit claims that a delta exists, we must exhibit the value of delta. We use the value for delta that we found in our preliminary work above, but based on the new second epsilon. Therefore, this delta is always defined, as \epsilon_2 is never larger than 72. Since \epsilon_2 >0, then we also have \delta >0. The definition does place a restriction on what values are appropriate for delta (delta must be positive), and here we note that we have chosen a value of delta that conforms to the restriction. The result is not real obvious, but can be seen as follows. Inside the square root expressions above, when subtracting from 25, the square root will be slightly smaller than 5, so the first delta candidate is positive. When adding to 25, the square root in the second candidate will be slightly larger than 5, so the second delta candidate is also positive. Therefore, their minimum is also positive. Now, for every x, the expression 0 < |x-c| < \delta implies This is the next part of the wording from the definition of the limit. -\delta < x-c < \delta -5+\sqrt{25-\dfrac{\epsilon_2}{3}} < x-5 < -5+\sqrt{25+\dfrac{\epsilon_2}{3}} When we have two candidates for delta, we need to expand the absolute value inequality so we can use both of them. Then we replace the values of c and delta by the specific values for this problem. From here on, we will be basically following the steps from our preliminary work, but in reverse order. \sqrt{25-\dfrac{\epsilon_2}{3}} < x < \sqrt{25+\dfrac{\epsilon_2}{3}} 25-\dfrac{\epsilon_2}{3} < x^2 < 25+\dfrac{\epsilon_2}{3} 75-\epsilon_2 < 3x^2 < 75+\epsilon_2$$-\epsilon_2 < 3x^2-75 < \epsilon_2$ We added 5 to each expression, then squared each expression, then multiplied each by 3, then subtracted 75. $|3x^2-75| < \epsilon_2 \le \epsilon$ Now we recognize that the two ends of our inequality are opposites of each other, so we can write the result as a single absolute value inequality. Furthermore, $\epsilon_2$ is always less than or equal to the original epsilon, by the definition of $\epsilon_2$. $|(3x^2-1)-74| < \epsilon$ Then we rewrite our expression so that the original function and its limit are clearly visible. Therefore,   $\lim\limits_{x\to 5} (3x^2-1)=74$. Having reached the final statement that   $|f(x)-L| < \epsilon$,   we have finished demonstrating the items required by the definition of the limit, and therefore we have our result. Q.E.D. Our proof is complete.