It's been a while since I've formally done any sort of calculus. Before moving on to more advanced topics in math, it's probably worth doing a minor review of the topic.
Starting with the basics, limits, we will work out a simple problem.
Find the tangent line of: $f(x) = -x^2 + 6\,\!$ where $x = 2$.
I guess, it may help to first define what a tangent line is and also a secant line since we will be using them both:
Tangent Line - A straight line is considered a tangent line of a curve if it goes through the point $(x, f(x))$.
Secant Line - A line is considered a secant line of a curve if it intersects two points on that curve. Assuming one of these points is $(x, f(x)$, these two points brought together, will give you the tangent line.
Now, in order to get a [tangent] line, we are going to need a slope. And, in order to get a slope, we're going to need two points. We already know how to get the first point (the point the tangent line must pass through) by solving for $(x, f(x))$ which, in this case, will be $(2, f(2))$ where, $f(2) = -2^2 + 6 = 2$, giving us our first point at $(2,2)$.
From here we can take our next point (for our secant line) at many different locations but, to keep things positive and simple, we'll use $x = 1$. This works out to $(1, f(1))$ where, $f(1) = -1^2 + 6 = 5$, giving us our second point at $(1, 5)$.
Taking the slope of these two points $\frac{5 - 2} {1 - 2}$, we now have a secant line (with a slope of $-3$) to work with. (Just to be clear, the line is $y = -3x + 8$)
As stated in the definitions, we can get the slope of the tangent line by bringing the points on the secant line closer together; more specifically, as the point $(1, 5)$ approaches the point $(2, 2)$, the closer the secant line gets to matching our tangent line. Using this knowledge, we can look at values of $x$ for our slope that are getting closer to $2$, such as $(1.5, f(1.5))$, $(1.8, f(1.8))$, and $(1.95, f(1.95))$. Solving for them we get, $-3.5$, $-3.8$, and $-3.95$ respectively.
Note: We can't use $2$ because we can't divide by zero. Also, it was an option to approach from the other side of the point using a point like $(3, f(3)) \Rightarrow (3, -3)$ and bring it closer to $(2, 2)$; this would give us the same result in this case but, that is not always the case as we will see later.
As you can see, the closer we get to the point $(2, 2)$, the closer we get to a slope of $-4$.
Using this slope, we can now express the tangent line as $y = -4x + 10$.
Now, expanding on this problem, $f(x) =$ some rate of change over period $x$. We can calculate this rate of change at different points using $\frac {f(x) - f(a)} {x - a}$, where $a =$ the point we are trying to reach. As $x$ approaches $a$, this point is called the Instantaneous Rate of Change.
For example, if we want to know the instantaneous rate of change at $1$ $(a = 1)$, we can follow these steps: $$\frac{f(x) - f(1)}{x - 1} \Rightarrow \frac{-x^2 + 6 - 5}{x - 1}$$ And, as $x$ approaches $1$ from both sides, we get closer and closer to $-2$.
Looking at the similarities in these two processes, you can infer that the instantaneous rate of change is also the same as the slope of the tangent line for a given function. Which leads us to the next point...
When you are trying to find the slope of a tangent line for a function, given a secant line, we only need to know one things, what point is $x$ trying to approach ($a$)? Now, we don't really need a secant line (explicitly, anyways) and without that, the arbitrary $a$ is not really needed either. All we really need to know is, what is the change in $f(x)$, represented by: $$(f(x + the distance from x) - f(x))$$ over the change in $x$, represented by: $$(x + the distance from x - x)$$The distance from $x$ can be represented by $\Delta x$. This looks like, $$\frac{f(x + \Delta x) - f(x)} {\Delta x}$$as $\Delta x$ approaches $0$.
In fact, we have a mathematical way of saying all of that, which is the limit: $$\lim \limits_{\Delta x \to 0}\frac{f(x + \Delta x) - f(x)}{\Delta x}$$This is also the mathematical definition of a derivative.
Now, the next calculus post will talk about derivatives so, let me just finish off here with a few examples of limits.
There are a few topics in limits I want to cover before moving on to derivatives: 3 methods of computing a limit and limits that you cannot evaluate. One thing to also keep in mind is that the limit of a function is NOT telling you what a function is at a point, rather what the function looks like as it approaches a point. This is important to understand as we move on.
In this first example, we will talk about the 3 methods of computing a limit. The three methods are: the substitution, the factoring and the conjugate method. The substitution method speaks for itself (substitute the variables and solve) so, I will combine it with the factoring method. Let's look at the limit: $$\lim \limits_{x \to 4}\frac{x^2 - 16}{x - 4}$$As you can see, we can't just plug in 4 for x so, we must first do some basic factoring: $$\lim \limits_{x \to 4}\frac{x^2 - 16}{x - 4}\Rightarrow \lim \limits_{x \to 4}\frac{(x + 4)(x - 4)}{x - 4}\Rightarrow \lim \limits_{x \to 4}x + 4$$Now, we can just substitute and see that: $$\lim \limits_{x \to 4}x + 4 = 8\Rightarrow \lim \limits_{x \to 4}\frac{x^2 - 16}{x - 4} = 8$$ When we can't substitute right away or factor, we turn to the conjugate method. Let's look at the limit: $$\lim \limits_{x \to 25}\frac{\sqrt{x} - 5}{x - 25}$$By multiplying by the conjugate we end up with: $$\lim \limits_{x \to 25}\frac{\sqrt{x} - 5}{x - 25} * \frac{\sqrt{x} + 5}{\sqrt{x} + 5}\Rightarrow \lim \limits_{x \to 25}\frac{(\sqrt{x} - 5)(\sqrt{x} + 5)}{(x - 25)(\sqrt{x} + 5)}\Rightarrow \lim \limits_{x \to 25}\frac{x - 25}{(x - 25)(\sqrt{x} + 5)}$$ $$\Rightarrow \lim \limits_{x \to 25}\frac{1}{\sqrt{x} + 5}$$And again, we can just substitute from here and see that: $$\lim \limits_{x \to 25}\frac{1}{\sqrt{x} + 5} = \frac{1}{10}\Rightarrow \lim \limits_{x \to 25}\frac{\sqrt{x} - 5}{x - 25} = \frac{1}{10}$$Finally, limits that do not exist. Let's look at the limit:$$\lim \limits_{x \to 0}g(x)$$where, $$when x < 0, g(x) = x and when x \geq 0, g(x) = x + 1$$Evaluating the limit for this is a little different than usual. In this case we will want to evaluate the limit from each side of $x$. When we want to take the limit from the left side, we denote it with a '-' and, when we want to take the limit from the right side, we use a '+'. For example:$$\lim \limits_{x \to 0^-}g(x) or \lim \limits_{x \to 0^+}g(x)$$These limits can be easily evaluated to $0$ from the left and $1$ from the right but, what about the standard limit?$$\lim \limits_{x \to 0}g(x)$$This limit does not exist (DNE) because the function is not continuous (it breaks at 0 in this case). One more example of this to try and make it clear.
Depending on the function and what value we're approaching, what the limit evaluates to can be quite different. Let's look at the limit:$$\lim \limits_{x \to 0}\frac{1}{x}$$In this example, both the left and and right hand limits exist, respectively giving us $-\infty$ and $\infty$. However taking the normal limit (from both sides) does not work. We see that:$$\lim \limits_{x \to 0}\frac{1}{x} = DNE$$This is again because, the function is not continuous. Graphing out these functions should help make this concept clear.
There are more to limits than what I've explained here but, hopefully this was a solid review. Also note that there are more methods available to help solve limits, such as L'Hôpital's rule, that you should check out when you have the chance. On my next calculus post, I will build on these concepts to review derivatives and some related concepts.
Here is a site for an online graphing calculator.
Any questions, comments and/or suggestions are appreciated.
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