Jim was watching his girlfriend run in a track meet. She was in the lead, and starting to pull away from the rest of the pack. Jim recognized a great photo opportunity, and snapped a great shot of Becca just as she rounded the corner and entered the homestretch.
Later, they discussed the race over a victory ice cream, as they admired the photo. "You were really moving, Becca," Jim noted.
"I felt like I was flying!" Becca replied.
"I wonder how fast you were running at the exact time I took the photo?" Jim mused.
"That's easy!" Becca said. "Just take the distance of the race, and divide it by the time it took me to run. Here, hand me your phone, I'll run it through your calculator app, what was my time?"
"Hold on, Becca," Jim interjected. "I don't think that will work. You weren't running the same speed the entire race, so dividing your total distance by your total time isn't much more than an educated guess of your speed the instant I took the pic. Maybe we could use the official race recording, it will be timestamped, and we could reference the track distance markers..."
"Oh, come on Jim! There is no way we can know what my speed was at that instant!" Becca countered. "Any calculation we come up with is going to be no more than an approximation!"
Is Becca right?
Watch This
James Sousa: Introduction to the Derivative
Guidance
The discovery of calculus was motivated by two fundamental geometric problems: finding the tangent line to a curve and finding the area of a planar region. In this section, we will show that these two problems are related to a deeper concept of calculus known as the limit of a function.
The Two Fundamental Problems of Calculus that Lead to its Discovery:


The portion of calculus that deals with the tangent problem is called differential calculus and the portion that deals with the area problem is called integral calculus. In order to solve those two problems, we need to have a more precise understanding of what a tangent line is and what is meant by the area under a curve. Both of these issues require us to understand a deeper concept, the limit of a function.
Tangent Lines and Limits
From your studies in geometry, you know that the tangent line is a line that intersects the circle at one point. However, this definition is not precise when we try to apply it to other kinds of curves. For example, as the figure below shows, one can draw a tangent line to a curve yet it cuts the curve at more than one point.
So we need to renew our concept of the tangent line and extend it to apply to curves other than circles. To do so, consider point P on the curve in the figure below. If point Q is any other point on the curve that is different from P, the line that passes through P and Q is called the secant line. Imagine if we move point Q along the curve toward point P, the secant line in this case will “rotate” toward a limiting position at point P. Eventually, the secant line will become a tangent line at point P, as the figure below shows. This is a new concept of the tangent line, where the general notion of a tangent line leads to the concept of limit. We will deal with the tangent line in more detail in lesson 8.3.
Area as a Limit
Suppose we are interested in finding the area under the curve of a function on the interval [a, b]. For example, consider function f(x) = (x  2)^{3} + 1 (the first figure below). Let’s say we want to approximate the area under the curve from x = 1 to x = 3. One way to do it is to inscribe rectangles of equal widths on the interval [1, 3] under the curve and then add the areas of these rectangles (the second figure below). Intuition tells us that if we repeat the process using more and more rectangles to fill the gaps under the curve, our approximation will approach the exact value of the area under the curve. So, the limiting value of this approximation is the exact value of the area under the curve. If we denote the width of each rectangle by ∆ x and the value of the area under the curve by A, then as ∆ x approaches zero (the widths of the rectangles get thinner and thinner, and thus less and less gaps), then the area A under the curve will reach an exact value.
What we have seen so far is that the concepts of tangent line and area rest on the notion of limit. In the next sections, we will explore those concepts in more details and show how the limit can help us calculate the rate of change of a given quantity. First, however, we introduce some useful notations.
Definition of a Limit (an informal view)


Example A
Make a conjecture about the value of the limit of \begin{align*}\lim_{x \rightarrow 0} \frac{3x} {\sqrt{x + 1}  1}\end{align*}
Solution:
Notice that the function \begin{align*}f(x) = \frac{3x} {\sqrt{x + 1} 1}\end{align*}
x  0.01  0.001  0.0001  0.00001  0  0.00001  0.0001  0.001  0.01 

f(x)  5.984962  5.9985  5.99985  5.999985  Undefined  6.000015  6.00015  6.0015  6.014963 
Another way of seeing this is to graph f(x) (shown below). Notice that the xvalues approach 0 from the left side and from the right side. In both cases, the values of f(x) appear to get closer and closer to 6.
Hence, again our conjecture is that \begin{align*}\lim_{x \rightarrow 0} \frac{3x} {\sqrt{x + 1}  1} = 6\end{align*}
Example B
Make a conjecture about the value of the limit \begin{align*}\lim_{x \rightarrow 0} \frac{sin x} {x}\end{align*}
Solution:
The function here is not defined at x = 0. With the help of a computing utility, we can obtain the table below.
x  0.2  0.1  0.01  0  0.01  0.1  0.2 

f(x)  0.993347  0.998334  0.999983  Undefined  0.999983  0.998334  0.993347 
The data in the table suggest thats \begin{align*}\lim_{x \rightarrow 0} \frac{sin x} {x} = 1\end{align*}
Example C
Make a conjecture about the value of the limit \begin{align*}\lim_{x \rightarrow 0} \frac{1cos x} {x^2}\end{align*}
Solution:
Enter the expression into your graphing calculator, or use this excellent free one here: https://www.desmos.com/calculator
You should get an image like the one below:
It is clear from the graph that the limit is 1/2.
Concept question wrapup: Technically, Becca is correct. However, using calculus to find the limit of her average speed at shorter and shorter intervals around the time the pic was taken could give Jim an answer that would be very, very close, as accurate as the race timer itself anyway. 

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Guided Practice
1) Use a grapher to make a conjecture about the value of the limit \begin{align*}\lim_{x\rightarrow 1} \frac{ln x} {2x  2}\end{align*}
2) Use a grapher to make a conjecture about the value of the limit \begin{align*}\lim_{x\rightarrow 0} \frac{tan2x} {x}\end{align*}
3) Use limit notation to write "The limit of f(x) equals the cosine of x, as x approaches 2 from the right."
Answers
1) Using a graphing calc (or https://www.desmos.com/calculator), the limit of 1/2 is easily located:
2) Using a graphing tool:
The limit is 2
3) \begin{align*}\lim_{x\to 2^+} cos x\end{align*}
Explore More
Write using limit notation:
 Write the limit of \begin{align*}4x^3 + 3x^2  4x  1\end{align*}
4x3+3x2−4x−1 as \begin{align*}x \end{align*}x approaches \begin{align*}a\end{align*}a from the left.  Write the limit of \begin{align*}g(z)\end{align*}
g(z) as \begin{align*} z \end{align*}z approaches \begin{align*} a \end{align*}a from the left.  Write the limit of \begin{align*}g(y)\end{align*}
g(y) as \begin{align*} y \end{align*}y approaches \begin{align*} b \end{align*}b from the left.  Write the limit of \begin{align*}h(z)\end{align*}
h(z) as \begin{align*} z \end{align*}z approaches \begin{align*} 1 \end{align*}−1 from the right.  Write the limit of \begin{align*}h(y)\end{align*}
h(y) as \begin{align*} y \end{align*}y approaches \begin{align*} a \end{align*}a from the left.  Write the limit of \begin{align*}h(z)\end{align*}
h(z) as \begin{align*} z \end{align*}z approaches \begin{align*} a \end{align*}a .
Solve using a calculator to estimate the limit:
 \begin{align*}\lim_{x\to0}\frac{\sqrt{4x+2}  \sqrt{2}}{4x}\end{align*}
limx→0−4x+2−−−−−−−√−2√4x  \begin{align*}\lim_{x \to1}\frac{8x^2  14x  6}{2x  2}\end{align*}
limx→−1−8x2−14x−6−2x−2  \begin{align*}\lim_{x\to0}\sec (\cos x)\end{align*}
 \begin{align*}\lim_{x\to\frac{16}{5}}\frac{\frac{2}{2x + 2} \frac{5}{11}}{10x  32}\end{align*}
 \begin{align*}\lim_{x \to\frac{5}{2}} \frac{\sqrt{x + 5}  \sqrt{5}}{2x + 5}\end{align*}
 \begin{align*}\lim_{x\to 4}\frac{x^2 + 6x + 8}{x + 4}\end{align*}
 \begin{align*}\lim_{x \to\frac{13}{2}}\frac{\frac{5}{2x + 3}  \frac{1}{2}}{2x  13}\end{align*}
 \begin{align*}\lim_{x\to0} \cot (\sin x)\end{align*}
 \begin{align*}\lim_{x\to0}\frac{\sqrt{x + 3}  \sqrt{3}}{5x}\end{align*}
 \begin{align*}\lim_{x\to0} \tan (\cos x)\end{align*}
Write a formal definition for the following problems:
 \begin{align*}\lim_{y\to2} \tan (y) = L\end{align*}
 \begin{align*}\lim_{x\to1} f(x) = N\end{align*}
 \begin{align*}\lim_{y\to1} x^3 + 2x^2 + 2x + 4 = L\end{align*}
Answers for Explore More Problems
To view the Explore More answers, open this PDF file and look for section 8.1.