Limits, the First Step into Calculus

Limits, the First Step into Calculus

The concept of a limit is the central idea that underlies calculus and is the unifying mechanism that allows for differentials and integrals to be related. Calculus is used to model real-life phenomena in the language of mathematics. Anything that involves a rate of change, as the velocity of your car is the rate of change of distance with respect to the rate of change of time, is found using derivatives. Limits are the basis of the derivative, by finding the instantaneous rate of change.

Definition of a Limit

The limit is the behavior of a function as we approach a certain value. Let’s start by looking at a particular function

$$f(x) = x^2 + x – 6$$

for values near 2. We can use a table of values that gets really close to 2 from values less than 2, and another that gets really close to 2 from values greater than 2.


From the table, we can see that as x approaches 2, the value of f(x) approaches 0. It would appear from the chart, that if we let x get really close to 2 in either direction that f(x) becomes 0. This is the basic version of how we solve a limit. We use the English phrase “the limit of f(x) as x approaches 0 is equal to 0”.

The Limit of a Function: Definition

We say

$$\lim_{x\rightarrow a} f(x) = L$$


$$\text{“The limit of f(x) as x approaches a equals L”}$$

if when we make the value of x get arbitrarily close to a, the value of f(x) gets arbitrarily close to L.

Finding Limits by “Direct Injection”

If we are searching for a limit like

$$\lim_{x\rightarrow 5} x^2+x-10$$

we can do what is called “Direct Injection”, in other words we plug in the value of \(x=5\) into the function we are finding the limit of

$$\lim_{x\rightarrow 5} 5^2+5-10=20$$

then we have discovered that the limit is equal to 20.

Try this method on the following problem:


Find the limit

$$\lim_{x\rightarrow 1} \frac{x-1}{x^2-1}$$


If we try direct injection we have a problem:

$$\lim_{x\rightarrow 1} \frac{1-1}{1^2-1} = \frac{0}{0}$$

The problem is that we cannot ever divide by zero. This is an undefined function in mathematics and algebra. We need another method to figure out how to take this limit. We are allowed to manipulate the function algebraically as long as we do not break any math rules. Notice that the denominator is factorable

$$\lim_{x\rightarrow 1}\frac{x-1}{x^2-1} = \lim_{x\rightarrow 1} \frac{x-1}{(x-1)(x+1)}$$

Now we can see that (x-1) is found in the numerator and denominator. We can simplify the expression into

$$\lim_{x\rightarrow 1}\frac{1}{x+1}$$

And now if we do the direct injection of x=1 we get

$$\lim_{x\rightarrow 1} \frac{1}{x+1} = \frac{1}{2}$$

And we have discovered the limit!


The limit is the behavior of a function as the variable approaches a specific number. Limits can be found in numerous different ways, this post has shown you two specific methods to discover limits:

  1. Table of values: Take values getting really close to the value you are searching for and measure the behavior of f(x).
  2. Direct Injection: Try plugging in the value you are searching for directly into f(x), and if it fails, try manipulating the equation using standard algebra techniques.

Check back soon for more information on Limits and Calculus in general!

What is an Integral?

What is an Integral?

The integral is a method to find the area under a curve. It is formulated as a sum of many smaller areas that approximate the area of the curve, all added together to find the total area. We let the number of areas under the curve approach infinity so the approximation to the area becomes the actual area under the curve. This is called a Reimann Sum.

Integrals were created to find the area, but it was discovered that they are related to derivatives. This discovery leads to integrals being what is called an “antiderivative.” The fundamental theorem of calculus attaches the theory of derivatives and integrals together, forming what we consider modern calculus.

The Problem: How To Find Area Under A Curve?

Area has always been of interest to society. Ancient farmers needed ways to divide up the land, which ideally would be located adjacent to a river. Since rivers wind back and forth in almost unpredictable ways, ensuring that each farmer received the same area in which to grow crops, a method for determining areas was required.

In trying to solve for the area of anything, we have to ask ourselves ‘What is the meaning of area?’

If you have a square, or rectangle, or any shape with straight edges, the area is somewhat easy to calculate. But what about the farmers with the winding river? How do you choose the boundaries of a curving river to be a straight side? You would have to start taking approximations of the river using straight lines in order. I used straight lines in my crude drawing above to try and segregate the proposed farm sections. But you may be able to spot that my plots are not all the same area. Some farmers would complain!

For a rectangle, the area is found by multiplying the length and the width. The area of a triangle is half the base multiplied by the height. The area of a polygon can be discovered by compartmentalizing it into triangles and adding the areas of the triangles.

We have methods to solve the areas of straight lines. But what about curved lines? We need a precise definition of the area. Let us start with a general curve:

general curve graph

We have no way (yet) to calculate the area found underneath this curve. So to make a crude approximation, we will draw rectangles whose height is from the x-axis to the function, and whose width is chosen so there are 5 equal width rectangles.

crude approximation of integral

You can see the first few rectangles overestimate the area under the curve, while the last few underestimate it. But if we add up all the areas of the rectangles, we arrive at a simplistic approximation to the area under the curve.

$$\text{Area} = f(x_0)\times (x_1-x_0)+f(x_1)\times (x_2-x_1)+f(x_2)\times (x_3-x_2)+f(x_3)\times (x_4-x_3) + f(x_4)\times (x_5-x_4)$$

If we know that all the \(x_i\) points are the same distance apart, we can rename that distance to \(\Delta x\) where \(\Delta x_i = x_{i+1}-x_{i}\)

Then we can re-write our sum of all rectangles in summation notation.

$$ \text{Area}=\sum_{i=0}^{4} f(x_i) \Delta x_i $$

Now imagine if we have a lot more rectangles:

Hopefully, it is obvious to you that by using many more rectangles of smaller widths we have reduced the error in how much each rectangle overestimates or underestimates the height of the curve. If we continue this trend and let the number of rectangles approach infinity, and the width of the rectangles approaches zero, then the accuracy of the area under the curve becomes perfect. We write such a notion like this:

$$\text{Area}= \underset{n \to \infty}{\lim_{\Delta x \to 0}} \sum_{i=0}^{n} f(x_i)\Delta x$$

By letting the distance between the two points make the width of the rectangle go to zero, we let the number of rectangles approach infinity by using the limit. The result is a perfect representation of the area under a curve. We call this result The Reimann Sum and we give it a special name:

The Integral.


$$\text{Area}=\text{integral} = \underset{n \to \infty}{\lim_{\Delta x \to 0}} \sum_{i=0}^{n} f(x_i)\Delta x = \int_a^b f(x) dx$$

The symbol \(\int\) stands for the integral, \(a\) and \(b\) are called the bounds of integration, the \(dx\) stands for \(\Delta x\) and represents an infinitesimal amount of \(x\), and \(f(x)\) is the curve we are looking for the area under.

The Natural Logarithm Rules

The Natural Logarithm Rules

The natural logarithm, whose symbol is ln, is a useful tool in algebra and calculus to simplify complicated problems. In order to use the natural log, you will need to understand what ln is, what the rules for using ln are, and the useful properties of ln that you need to remember.

What is the natural logarithm?

The natural logarithm is a regular logarithm with the base e. Remember that e is a mathematical constant known as the natural exponent. We write the natural logarithm as ln.

$$\log_e (x) = \ln(x)$$

Since the ln is a log with the base of e we can actually think about it as the inverse function of e with a power.

$$\ln(e^x ) = x$$
$$e^{\ln(x)} = x $$

The natural exponent e shows up in many forms of mathematics from finance to differential equations to normal distributions. It is clear that the logarithm with a base of e would be a required inverse so as to help solve problems involving such exponents.

Properties of ln

  1. \(\ln(a)\) exists if and only if \(a>0\)The natural logarithm of a requires that a is a positive value. This is true of all logarithms. This is an important parameter to remember, as any logarithm of a negative number is undefined.
  2. \(\ln(0)\) is undefinedNotice how in property 1 that we define \(\ln(a)\) to exist if \(a > 0\). That is no mistake. The logarithm of zero is undefined.
  3. \(\ln(1)=0\)The natural logarithm of 1 is 0. This is a useful property to eliminate certain terms in an equation if you can show that the value in the natural logarithm is 1. It also serves as a divider between solutions of the natural log that are either positive or negative. \(\ln(a) < 0\) if \(0 < a < 1\) and on the other side \(\ln(a) > 0\) if \(a > 1\).
  4. \(\lim\limits_{a\rightarrow\infty} \ln(a)=\infty\)The limit as \(ln(a)\) as a approaches infinity is infinity. The natural logarithm is a monotonically increasing function, so the larger the input the larger the output.
  5. \(\ln(e)=1\)Since the base of the natural logarithm is the mathematical constant e, the natural log of e is then equal to 1.
  6. \(\ln(e^x)=x\)Since the natural logarithm is the inverse of the natural exponential, the natural log of e x becomes x.
  7. \(e^{\ln(x)}=x\)Similar to property 6, the natural exponential of the natural log of x is equal to x because they are inverse functions.

The Natural Logarithm Rules

There are 4 rules for logarithms that are applicable to the natural log. These
rules are excellent tools for solving problems with natural logarithms involved,
and as such warrant memorization.

  1. The Product Rule$$\ln(ab)=\ln(a)+\ln(b)$$

    If you are taking the natural log of two terms multiplied together, it is equivalent to taking the natural log of each term added together.

    Note 1: Remember property 1. The natural log of a negative value is undefined. This implies that both terms a and b from the product rule are required to be greater than zero.

    Note 2: This property holds true for multiple terms:

  2. The Quotient Rule$$\ln\left(\frac{a}{b}\right)=\ln(a)-\ln(b)$$

    If you take the natural log of one term divided by another, it is equivalent to the natural log of numerator minus the natural log of the denominator.

    Note 1: Remember property 1. The natural log of a negative value is undefined. This implies that both terms \(a\) and \(b\) from the quotient rule are required to be greater than zero.

  3. The Reciprocal Rule$$\ln\left(\frac{1}{x}\right)=-\ln(x)$$

    If you take the natural log of 1 divided by a number, it is equivalent to the negative natural log of that number.

  4. The Power Rule$$\ln(a^b)=b\ln(a)$$

    If you take the natural log of a term \(a\) with an exponent \(b\), it is equivalent to \(b\) times the natural log of \(a\).

It is of use to any student to be able to prove these 4 rules of natural logarithms. The observant student will see that the product rule can be proved easily using property 6 and 7, and some knowledge of exponents. The quotient, reciprocal, and power rule all follow from specific versions of the product rule. So if you are able to prove the product rule, the remaining three should be trivial.


  • The natural log ln is a logarithm with a base of the mathematical constant e. ie \(\ln=\log_e\)
  • The natural log ln is the inverse of e
  • The 4 rules of logs
    • $$\ln(ab)=\ln(a)+\ln(b)$$
    • $$\ln\left(\frac{a}{b}\right)=\ln(a)-\ln(b)$$
    • $$\ln\left(\frac{1}{x}\right)=-\ln(x)$$
    • $$\ln(a^b)=b\ln(a)$$

Navier-Stokes is hard?

I was recently asked the question; “How can the Navier-Stokes equations both describe our observable world and not be known to always have solutions in 3D?

For those that don’t know, the Navier-Stokes (NS) equations represent a mathematical model that describes fluid. It is derived from quantities in physics that are conserved. Those are:

  • Conservation of mass
  • Conservation of momentum
  • Conservation of energy

The Navier-Stokes equations are yet to be considered solved. There is a prize association called “The Millennium Prize Problems” which offers one million dollars to anyone that can solve one of the seven problems they have listed, and the existence and smoothness of a solution to the Navier-Stokes equations is on that list.

So the question I was asked seems completely reasonable. How can these equations describe physical phenomena with knowledge of a solution? The answer touches on an interesting, and modern, intersection in science.

We use a number of assumption to simplify a NS system in order to approximate a solution. These approximations are usually handled by a computational fluid dynamics software package. They are high powered suites that allow for customized physics simulations. The assumptions most often used to simplify the NS equations are things like the fluids being in-compressible, the turbulence affects are averaged, or that the fluid is Newtonian. Suffice to say, even with a high powered computational engine, more often then not a scaled down and simplified version of the NS equations are all that we can easily approximate.

So if the solutions to the fluid problems are all approximate, how can we know if they are describing real physics? To compare a NS system solution to an actual physical fluid flow, we would need to perform an experiment and collect data on such a flow. This results in flawed data, because the act of introducing measurement tools into fluid domains involves disrupting the fluid. The best measurements of fluid dynamics themselves are approximations.

So what we have ended up with is approximate mathematical approximations of fluids approximating approximate experiments. Its not hard to see why the student’s question needed clarification.

The answer is: Because it works.

That’s a pretty hand-waved explanation, that no student should accept. But for anyone interested, the derivation of Navier-Stokes is a fun derivation that follows from those conservation laws that were listed, and experiments can be found in open source publications that show how much effort was placed on collecting data with minimal interference. The interested party might be able discern that experimental data matches model solutions (reasonably well), and we can start to believe that Navier-Stokes delivers on the promise of predicting real world physics.

The Quadratic Equation

The Quadratic Equation

The quadratic equation is always the answer.

Time and time again,  while tutoring students,  I encounter a resistance to using  an  unfailing  tool  called  the  quadratic  equation.   This  simple  algebraic mathematical statement allows a student to find the roots of any clumsy second order polynomial with ease. I can only assume that, while the quadratic equation is drilled into memory, its useful nature is under-reported in high school mathematics.

For those who aren’t familiar:

\( ax^2 + bx + c = 0 \)                (1)

\( x=\frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \)                    (2)

If you can get an equation into the form of equation??, then you can use the Coefficients a, b, and c in order to solve for x in equation??. The ± in equation?? implies that there are two solutions for x (this being the result of solving a second order polynomial equation).  These two solutions represent the roots of equation??. The roots are the values of x that allow equation?? to equal 0 on the left hand side. Roots  are  of  interest  to  us,  because  they  are  a  quick  and easy  way  to determine factors of a quadratic (not to mention all the mathematical reasons to identify when a function is equal to zero).  Students are drilled on factoring. I rarely see a student that is unable to factor a simple quadratic, but I often encounter  those  that  hav problems  when  the  factors  fail  to  be  immediately obvious.  The quadratic equation is always the answer. I often despair that education relies entirely on problem singularity, a belief that there is only one correct way to approach a question.  When I see a student struggle  for minutes  to  try  and  factor  a  difficult  quadratic,  I  can’t  help  but wonder why they shy away from the quadratic equation.  Sometimes it is hyper- focus that keeps them from finding an alternate path forward,  but more and more I encounter fear.  Factoring is supposed to be an easy shortcut, while the quadratic equation has the confusing ± and the scary √ .  But a shortcut fails to be short when it takes time to muddle through possible variations.  Take, for example, the following quadratic:

 \( 10x^2 + 13x – 30 = 0 \)        (3)

Some  people  may  be  able  to  see  immediately  that  this  equation  can  be factored.  Others  might have enough experience with factoring to only fumble through one or two permutations before generating the factored result quickly. The  rest  of  us,  myself  included,  might  struggle  with  5  or  more  permutations before coming close to a factored result.  Yet the quadratic yields the answer in 4 quick lines:

\( x=\frac{-13 \pm \sqrt{169 + 1200}}{20} \)              (4)

\( x=\frac{-13 \pm \sqrt{1369}}{20} \)                     (5)

\( x=\frac{-13 \pm 37}{20} \)                           (6)

so \( x= \frac{6}{5} \) and \( x=\frac{-5}{2} \)         (7)


The quadratic equation is always the answer.

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