Problem: Let \(f(x)=x^2+3x\), Let \(a=1\). Find \(f^{\prime}(x)\).
Solution: Set up the difference quotient
\(\lim_{x\rightarrow a} \frac{f(x)-f(a)}{x-a}=\lim_{x\rightarrow a}\frac{x^2+3x-4}{x-1}=\lim_{x\rightarrow a}\frac{(x-1)(x+4)}{x-1}=\lim_{x\rightarrow a} x+4=1+4=5=f^{\prime}(1)\).

Remark 1.1.  The calculus is a mathematical system in which:

1. The basic elements are functions.
2. The basic concept is the concept of a limit of a function.
3. The algebraic binary operations of addition and multiplication of pairs of functions are used.
4. Major calculus operations are differentiation of a function and integration of a function.
5. Many properties of functions and operations on functions involve limits and continuity.

Remark 1.1 is not intended as a definition, but is intended as a partial description of the calculus. The system of real numbers is another mathematical system which is of utmost importance and is heavily used within the mathematical system called the calculus.
Each calculus student spent a great deal of time (in elementary and high school classes) learning basic properties which hold within the system of real numbers. The student can recall that it became very important (and very useful) for us to:

1. choose some particular straight line,
2. pick one particular point on this line and call it the origin, and

Definition.

Let \(f\) be a real function of a real variable \(x\); also let \(L\) and \(a\) be real numbers. Then
\(L\) \(=\) \(\lim_{}\) \(f(x)\) iff the following condition holds: For each open interval \(U\) centered at \(L\),
there is some open interval \(V\) centered at \(a\) such that if \(x\) is in \(V^{\ast}\), then \(f(x)\) is in \(U\).

Example 1.

Let \(y = f(x) = 3x + 4.\) Use Def __ to show that \(\lim_{} = 10.\)

We should review Definition __ before giving a solution for this example.

{Draw figure before the word solution.}

Solution. We let \(U\) be an arbitrary open interval centered at \(L=10\) with radius \(r>0\). So we know:

 

 

(1)                          If \(|f(x) – 10| <r\), then \(f(x)\) is in \(U\).

 

Now we express \(f(x)\) as \(3x+4\) and we have

 

\(\qquad\qquad |f(x)-10|= |(3x+4) – 10| = |3x – 6| = 3|x-2|\)

 

 

or more briefly

 

 

(2)                          \(|f(x) – 10| = 3|x-2|.\)

 

 

Because of (2), we let \(V\) be the open interval centered at \(a=2\) with radius \(\frac{r}{3}\). Now we let \(x\) be a point in \(V^{\ast}\). Then \(x\) is in \(V\) so that \(|x-2|<\frac{r}{3}\). Therefore, by (2), we have

 

 

\(\qquad\qquad|f(x)-10| = 3|x-2| < 3(r/3) = r\)

 

or briefly,

 

(3)                         \( |f(x)-10| < r\)     if \(x\) is in \(V^{\ast}\).

 

 

Now by (3) and (1), we have shown that if \(x\) is in \(V^{\ast}\), then \(f(x)\) is in \(U\). Therefore, \(\displaystyle 10 = \lim_{x\rightarrow 2} f(x)\)._