The history of calculus is interconnected with the history of philosophy and scientific discover. In the 16th Century, French philosopher Rene Descartes combined principles of algebra and geometry by mapping lines and objects using horizontal and vertical planes. By applying numbers to different points on geometric shapes, Descartes created what we now refer to as “graphing.” In doing so, Descartes was able to represent geometric shapes using algebraic equations.

Over 100 years later, two famous European thinkers used Descartes mathematical advancements to discover calculus. Sir Isaac Newton and Gottfried Wilhelm Leibnitz are generally credited with the discovery, though there are disputes over who established the basic function and purpose of calculus first. Newton is most famous for formulating laws of motion and gravity. As a direct result of his work with universal gravitation, Newton stumbled upon a quandary that would lead him to the discovery of calculus: how can the speed of a falling object be measured at a singular point in time?

While Newton was trying to figure this out, Leibniz was also trying to figure out how to understand and define quantities that are constantly changing. The result of their combined work was the creation of derivatives, which are a particular kind of slope. Slopes were only understood as a result of Descartes discovery of graphing, and thus, after hundreds of years of trials and theories, calculus was born.

At its most basic form, calculus is the study of continuously changing quantities. One of the most fundamentally important elements of calculus are limits. A **limit** is the value that a function approaches as an input reaches another value. That definition may seem a little daunting, so let’s break it down.

The precise definition of a limit is given and it is shown through examples why the definition is needed. The formal definition of the limit is what allows us to prove results so that more analytic techniques for evaluating limits can be accomplished.

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## Why is Precision Needed For Limits?

You may have already seen examples of estimating limits numerically and graphically. Each of these approaches produces an estimate of the limit, but it is the formal definition of the limit that allows us to prove results so that more analytic techniques for evaluating limits can be accomplished. In summary, a three-pronged approach to solving limits is often:

(i) numerical approach by constructing tables of values

(ii) graphical approach by sketching a graph by hand or using technology

(iii) analytic approach by using algebra or calculus.

We will use a guessing method to show why the formal definition of a limit is a necessity.

**Example**. Use a table to guess the values of $$ L=\lim_{x\to 0}\frac{2\sqrt{x+1}-x-2}{x^2}. $$

**Solution**. From the table $$ \begin{array}{c|cccccc} x & -0.1 & -0.01 & -0.001 & 0 & 0.001 & 0.005 \\ \hline f(x) & -0.2633 & -0.2513 & -0.2501 & \text{undefined} & -0.2499 & -0.2494 \end{array} $$ The number $L$ is suggested to be $-0.25.$ Interestingly, if you try ${x=0.0000001}$ just to make sure you have taken numbers close enough to 0, you may find that the calculator gives the value 0. Does this mean that the limit is 0? No, the calculator may give you a false answer because when $x$ is small enough (like $0.0000001$) then $2\sqrt{x+1}-x-2$ seems like 0. But in fact $f(0.0000001)$ is not equal to 0. The point is, using technology to verify a computation can lead to misunderstanding; and in fact, a formal definition of a limit is needed.

Using the formal definition of a limit, we can prove what the value of the limit is without any doubt. This type of proof is usually called an epsilon-delta proof since the formal definition is usually stated with the greek letters $\epsilon $ (epsilon) and $\delta $ (delta).

**Example**. Use tables of values to find the limit $$ \lim_{x\to 0}\left(x^3+\frac{\cos 5x}{10,000}\right). $$

**Solution**. As before, we construct a table of values. $$ \begin{array}{c|ccccc} x & 1 & 0.5 & 0.1 & 0.05 & 0.01 \\ \hline f(x) & 1.000028 & 0.124920 & 0.001088 & 0.000222 & 0.000101 \end{array} $$ From the table it appears that $$ \lim_{x\to 0}\left(x^3+\frac{\cos 5x}{10,000}\right)=0. $$ However, if we persevere with smaller values of $x,$the next table suggests $$ \lim_{x\to 0}\left(x^3+\frac{\cos 5x}{10,000}\right) =0.0001000=\frac{1}{10,0000} $$ $$ \begin{array}{c|cc} x & 0.005 & 0.001 \\ \hline f(x) & 1.00010009 & 0.00010000 \end{array} $$ In fact, $$

\lim_{x\to 0}\left(x^3+\frac{\cos 5x}{10,000}\right) =\frac{1}{10,0000} $$ which is easily proven once the formal limit definition is used to prove some interesting limit rules and continuity is discussed.

## The Precise Definition of a Limit

**Definition**. Suppose that the domain of $f$ contains points $x$ arbitrarily close to c but different from $c$. Then $$ \lim_{x\to c}f(x)=L $$ means, for all $\epsilon >0$ there exists $\delta >0$, such that $$ |x-c|<\delta \Rightarrow |f(x)-L|<\epsilon $$ for any $x\neq c$ in the domain of $f.$ Concisely, a limit is used to describe the behavior of a function near a point but not at the point. The function need not even be defined at the point. If it is defined there, the value of the function at the point does not affect the limit. Intuitively, $$ \lim_{x\to c}f(x)=L $$ means we can make $f(x)$ as close to $L$ as we wish by taking any $x$ sufficiently close to, but different from $c.$

## Epsilon-Delta Argument

**Example**. Use an epsilon-delta argument to show that $$ \lim_{x\rightarrow -2} (7x+12)=-2. $$

**Solution**. Let $\epsilon $ be any real number greater than zero. We need to find a $\delta >0$ so that, $$ |x-(-2)|<\delta \Rightarrow |7x+12-(-2)|<\epsilon . $$ To do so note that, $ |(7x+12)-(-2)| =7|x+2|.$ We find that $\delta =\epsilon /7$ is acceptable because, if $|x-(-2)|<\delta $ then $$ |7x+12-(-2)| =7|x+2| <7\delta =7\left(\frac{\epsilon }{7}\right) =\epsilon $$ as needed.

**Example**. Use an epsilon-delta argument to show that $$ \lim_{x\rightarrow \frac{-1}{3}} \left(-\frac{2}{3}x^2-\frac{4}{5}x-\frac{2}{3}\right)=\frac{-64}{135}.$$

**Solution**. Let $\epsilon $ be any real number greater than zero. We need to find a $\delta >0$ such that, $$\left|x+\frac{1}{3}\right|<\delta \Rightarrow \left|\left(-\frac{2}{3}x^2-\frac{4}{5}x-\frac{2}{3}\right)-\left(\frac{-64}{135}\right)\right|<\epsilon .$$ To do so note that,

\begin{align} & \left|\left(-\frac{2}{3}x^2-\frac{4}{5}x -\frac{2}{3}\right)-\left(\frac{-64}{135}\right)\right| =\left|-\frac{26}{135}-\frac{4 x}{5}-\frac{2 x^2}{3}\right| \\ & \qquad =\frac{2}{3}\left|\frac{13}{45}+\frac{6 x}{5}+x^2\right| =\frac{2}{3}\left| x+\frac{13}{15}\right| \left|x+\frac{1}{3}\right| \end{align} which is useful because $\left|x+\frac{1}{3}\right|$ is small, say $\left|x+\frac{1}{3}\right|<1$ and thus,

\begin{align*} \left|x+\frac{1}{3}\right| < 1 & \Longleftrightarrow -1 < x+\frac{1}{3} < 1 \Longleftrightarrow \frac{-4}{3} < x < \frac{2}{3} \\ & \Longleftrightarrow \frac{-7}{15} < x+\frac{13}{15} \frac{23}{15} \Rightarrow \left|x+\frac{13}{15}\right| < \frac{23}{15} \end{align*}We find that $$ \delta =\min \left\{1,\frac{\epsilon }{\left(\frac{46}{45}\right)}\right\}$$ is acceptable because, if $$ \left|x+\frac{1}{3}\right|<\delta $$ then \begin{align} \left|\left(-\frac{2}{3}x^2-\frac{4}{5}x-\frac{2}{3}\right)-\left(\frac{-64}{135}\right)\right| & =\frac{2}{3}\left| x+\frac{13}{15}\right| \left|x+\frac{1}{3}\right| \\ & <\left(\frac{2}{3}\right)\left(\frac{23}{15}\right)\delta \\ & =\left(\frac{46}{15}\right)\frac{\epsilon }{\left(\frac{46}{15}\right)} \\& =\epsilon \end{align}

as required.