The concept of limits is one that builds the backbone of calculus. Most of the tedious, more important parts of calculus are defined and built off of the concept of limits, so in order to understand derivatives and integrals, understanding the concept of limits must first be shown.
Imagine if you’re across the room from a heater and you wonder how hot the heater actually is. When you’re across the room, you can barely feel any of the heat coming from the heater, so you begin to inch closer and closer toward the heater. As you approach the heater, you begin to feel the air warm around and you feel the heat much more strongly. As you get closer, right next to the heater, you feel an overwhelming pulse of heat engulfing you.
While you could touch the heater to see how warm it is, you may burn yourself since the heater is too hot to touch, so you hold your hand directly over the heater to estimate how warm it is.
All in all, to estimate the temperature of the heater, you approached the machine and noticed the change, and then once you got near it, you got extremely close to the heater and were able to estimate how hot it actually was on the surface of the heating unit.
How does this relate to limits? Well, in essence, a limit follows the same procedure as how you determined the temperature of the heater. All a limit does it follow a function toward a point of the interest (the heater itself), and you get oh-so very close to the point to estimate its value using the behavior of the function (holding your hand directly over the heater).
A limit evaluates the behavior of a function at a certain point without actually being on that point.
Why is this needed? Besides making the life of college students harder, limits allow us to get an insight on how a function behaves, whether that is at a specific point of interest or the end behavior of the graph (this is evaluated with limits involving infinity)
Why is this needed? Because sometimes in calculus, we need to assess how a function behaves at a given point, even when the function doesn't have a value at said point.
Assume you want to find the limit of any function, let’s call it f(x), at some specific point called a. We can write this limit as the following:
You would read this out loud as “The limit as x approaches a of f(x)”. X is our independent variable, a is the point of interest on the x-axis, and f(x) is the function we’re evaluating the limit on. It seems a bit hard to understand in general, but with examples, the concepts should be very apparent.
Sometimes you may see limits written as this:
...or like this:
These signs denote the type of limit being taken.
If you see a+, then the limit is called a right-handed limit, a limit that you evaluate approaching the point of interest, a, from the right side of the function.
If you see a-, then the limit is called a left-handed limit, a limit that you evaluate approaching the point of interest, a, from the left side of the function.
If you see a (with no sign present), then the limit is called an absolute limit, a limit that evaluates both the right and left side of the function as you approach the point of interest, a.
How to solve for limits
Evaluating limits is ultimately straightforward and tends to rely on numerous algebraic techniques you utilized in previous math courses. I will refresh you on these topics here while applying them to limits.
Direct evaluation / substitution
This is the simplest method to evaluate a limit, and it’s ultimately the most effective. Going back to the example mentioned above:
We can evaluate the limit of this function at a by just substituting in a for x in every part of the function, ultimately stating that:
Let's see how this would work in action!
If we were given the following prompt...
Question: Evaluate the limit of the below function at x = 3.
...we could set up solving for the limit at x = 3 like so:
To evaluate this limit with direct substitution, all we'd need to do is substitute 3 for x like so...
...and then solve to arrive at our limit of "56 2/3"!
What this means is that when you approach x = 3 from the left and right side of our function f(x), it approaches a y value of 56 2/3!
See how easy that was? That's why...
Direct evaluation / substitution should always be your first go-to when evaluating a limit! It’s the easiest and takes the least amount of time.
However, sometimes we won't be able to directly plug in an x value to evaluate a limit. In those situations, we'll need to simplify our function down to the point where we can do so! Let's dig into a couple of those situations below.
Limits that involve infinity
Infinity can be a confusing concept to grasp. That's why when working with limits, it's crucial to zone in on whether the numerator or denominator has the highest power of x.
When working with infinity in a limit, first assess if the highest power of x occurs in the numerator or denominator. That side of the fraction will grow to infinity the quickest.
Still confused? Don't worry, keep reading and it'll start making sense.
Highest power of x in denominator
Let's consider the following identity:
Translated to English: When you divide 1 by infinity...
...it will ultimately equal 0. That's because 1 is so small compared to infinity, that it's practically 0!
When the highest power of x occurs in the denominator, it is bottom-heavy and shrinks quickly. Therefore, the limit goes to 0.
Highest power of x in numerator
Now let's consider this identity (basically the opposite of the last one):
Translated to English: Infinity over 1 (or for that matter, any number less than infinity)...
...will ultimately equal infinity. That's because 1 (and any number less than infinity) is so much smaller than infinity, that the infinity will essentially overpower it.
When the highest power of x occurs in the numerator, it is top-heavy and grows quickly. Therefore, the limit goes to infinity.
Numerator and denominator have same power of x
What if we had the following function?
In this situation, we're able to identify that both the numerator and the denominator have the same highest power of x.
Therefore, we need to only zone in on these polynomials and discard the rest.
Think about it this way:
∞2 is so much larger than ∞, that it pretty much makes ∞ obsolete. Likewise, ∞3 is so much larger than ∞2, that it makes ∞2 obsolete. This is why when working with limits going to infinity, we only need to focus on the highest power of x!
From here, we can go ahead and cancel out the x2...
...to arrive at our final limit of 5!
When the numerator and denominator both have the same highest power of x, the limit is the ratio of the coefficients of the highest power of x!
When evaluating some limits that don't deal with infinity, you may run into issues with indeterminate forms.
Indeterminate form essentially means an unknown value that's not solvable mathematically. The most common that we'll work with is 0/0.
When faced with indeterminate forms with polynomials, you can attempt to factor the expression and potentially cancel out terms to simplify the expression. Let's see this in action in the below example:
Question: Evaluate the limit of the below function at x = 3.
Remember: always try direct substitution first and see what happens!
Darn! Unfortunately it doesn’t work out, but since we got 0/0, an indeterminate form, our work is not done yet! We must continue! Let’s try factoring the expression in the numerator, (the denominator is already factored down as much as it can be) like I suggested earlier and see if anything happens…
Interesting! To factor the numerator, all I did was ask what numbers multiply to -6...
...and add to -1...
...resulting in -3 and 2, respectively.
Once we have this factored out, notice that we have an (x - 3) term that we can cancel out readily. This term is actually what caused the indeterminate form to happen in the first place, so getting rid of this is key!
It caused the indeterminate form since both (x-3) terms become 0 when you plug in 3 to directly determine the limit, yielding the 0/0 indeterminate form. Let’s cancel out the term and evaluate the limit with direct substitution.
Now let's apply direct substitution and plug in 3 for x...
...to arrive at our limit of 5!
If you’re dealing with a ratio consisting of polynomials (also known as rational functions), try factoring and see how that goes.
This process is most useful when you’re evaluating limits with square roots or other radical expressions. All you do is multiply both parts of the expression by the conjugate of the expression. But… what’s a conjugate?
A conjugate is an expression with two values and the sign in the middle flipped.
This enables you to better simplify the expression. Let's check out this example below:
Question: Evaluate the limit of the below function at x = 16.
Like always, let’s try evaluating this limit with direct substitution first.
Darn! We got another indeterminate form, 0/0. Since the numerator has a radical in it...
...let's try rationalizing it by multiplying the numerator and denominator by the conjugate...
...which is simply multiplying by 1 (because the same expression over the same expression equals 1).
Remember, when you choose the conjugate, you keep the sign of the first part of the expression the same, but flip the sign of the second part. In other words, we keep the √(x) the same, but change the -4 to a +4.
Now let’s distribute the expression (FOIL) in the numerator and multiply the entire expression out. Hold off on the denominator for now. It will save us some time.
From here, we can easily factor out the (x - 16).
Now, apply direct substitution to arrive at our limit of 1/8!