We’ll look at this thing that we already know from a slightly different perspective (or maybe just a specific perspective), and we’ll discover a really important and helpful result from it!
This is an important thing to notice: we can change our function by changing the value of the function at \(x=a\) without changing the value of the limit as \(x\to a\text{.}\)
Theorem1.3.3.Limits of (Slightly) Different Functions.
If \(f(x)\) and \(g(x)\) are two functions defined at \(x\)-values around \(a\) (but maybe not at \(x=a\) itself) with \(f(x) = g(x)\) for the \(x\)-values around \(a\) but with \(f(a)\neq g(a)\) then \(\displaystyle \lim_{x\to a}f(x) = \lim_{x\to a}g(x)\text{,}\) if the limits exist.
Why will this be helpful? At the end of Section 1.2 we found that some functions (polynomials) are great: the limit of these functions is the same as the function value at the point (Theorem 1.2.7). This is a special case, and many functions won’t be so nice to work with. But maybe we could use Theorem 1.3.3 to swap out an annoying-to-work-with function for a nice-to-work-with function!
SubsectionA First Introduction to Indeterminate Forms
So before we begin applying this result, we will focus on a situation where we need it. We’re going to do something strange: define a situation before we experience it.
We say that a limit has an indeterminate form if the general structure of the limit could take on any different value, or not exist, depending on the specific circumstances.
For instance, if \(\displaystyle \lim_{x\to a} f(x) = 0\) and \(\displaystyle \lim_{x\to a} g(x) = 0\text{,}\) then we say that the limit \(\displaystyle \lim_{x\to a}\left(\frac{f(x)}{g(x)}\right)\) has an indeterminate form. We typically denote this using the informal symbol \(\frac{0}{0}\text{,}\) as in:
We’re going to see in this next activity that this kind of \(\dfrac{0}{0}\) form of a limit can actually lead to very different behavior: we call it indeterminate because we cannot determine, based solely on the form \(\dfrac{0}{0}\text{,}\) what the limit is or even if it will exist.
Now we want to find a new function that is equivalent to \(f(x) = \dfrac{x^2-7x+12}{x-3}\) for all \(x\)-values other than \(x=3\text{.}\) Try factoring the numerator, \(x^2-7x+12\text{.}\) What do you notice?
This function is equivalent to \(f(x) = \dfrac{x^2-7x+12}{x-3}\) except at \(x=3\text{.}\) The difference is that this function has an actual function output at \(x=3\text{,}\) while \(f(x)\) doesn’t. Evaluate the limit as \(x\to 3\) for your new function.
Now we want a new function that is equivalent to \(g(x)=\dfrac{\sqrt{x^2+3}-2}{x^2-5x+4}\) for all \(x\)-values other than \(x=1\text{.}\) Try multiplying the numerator and the denominator by \((\sqrt{x^2+3}+2)\text{.}\) We’ll call this the “conjugate” of the numerator.
This function is equivalent to \(g(x) = \dfrac{\sqrt{x^2+3}-2}{x^2-5x+4}\) except at \(x=1\text{.}\) The difference is that this function has an actual function output at \(x=1\text{,}\) while \(g(x)\) doesn’t. Evaluate the limit as \(x\to 1\) for your new function.
Again, we want a new function that is equivalent to \(h(x)=\dfrac{3-\frac{3}{x+3}}{x^2+2x}\) for all \(x\)-values other than \(x=-2\text{.}\) Try completing the subtraction in the numerator, \(3-\frac{3}{x+3}\text{,}\) using “common denominators.”
For the final time, we’ve found a function that is equivalent to \(h(x) =\dfrac{3-\frac{3}{x+3}}{x^2+2x}\) except at \(x=-2\text{.}\) The difference is that this function has an actual function output at \(x=-2\text{,}\) while \(h(x)\) doesn’t. Evaluate the limit as \(x\to -2\) for your new function.
In each of the previous limits, we ended up finding a factor that was shared in the numerator and denominator to cancel. Think back to each example and the factor you found. Why is it clear that these must have been the factors we found to cancel?
Let’s say we have some new function \(f(x)\) where \(\displaystyle \lim_{x\to 5} f(x) \stackrel{?}{\to} \frac{0}{0}\text{.}\) You know, based on these examples, that you’re going to apply some algebra trick to rewrite your function, factor, and cancel. Can you predict what you will end up looking for to cancel in the numerator and denominator? Why?
This is great: we’re applying Theorem 1.3.3 because in each algebraic manipulation, we change the domain of the function by removing some factor from the denominator!
These three algebra tricks are all we’ll look at for now. In reality, there are plenty of little tricky manipulations we can use to slightly change functions, but if we focused on trying to build one for every situation we could run into, we’d spend the rest of this text just outlining different algebra tricks for different situations.
Combine fractions with common denominators: This works well when we have some subtraction with fractions inside of a numerator or denominator of another fraction.
We’ve seen some nice examples above where we were able to use some algebra to manipulate functions in such was as to force some shared factor in the numerator and denominator into revealing itself. From there, we were able to apply Theorem 1.3.3 and swap out our problematic function with a new one, knowing that the limit would be the same.
This happens a lot, and we’ll investigate some more of those types of limits in Section 4.7. For now, though, let’s look at a very famous limit and reason our way through the indeterminate form.
Now let’s think about the meaning of \(\sin(\theta)\) and even \(\theta\) in general. In this text, we will often use Greek letters, like \(\theta\text{,}\) to represent angles. In general, these angles will be measured in radians (unless otherwise specified). So what does the sine function do or tell us? What is a radian?
On the unit circle, if we plot some point at an angle of \(\theta\text{,}\) then the coordinates of that point can be represented with trig functions! Which ones?
The length of the curve defining a unit circle is \(2\pi\text{.}\) This also corresponds to the angle we would use to represent moving all the way around the circle. What must the length of the portion of the circle be up to some point at an angle \(\theta\text{?}\)
Explain to yourself, until you are absolutely certain, why the two lengths must be the same in the limit as \(\theta\to 0\text{.}\) What does this mean about \(\displaystyle \lim_{\theta\to 0}\frac{\sin(\theta)}{\theta} \text{?}\)
When you were evaluating the limit, you likely “cancelled” a factor of \((x-3)\) from the numerator and denominator. Why might you have known, before factoring anything, that \((x-3)\) would be a factor shared in the numerator and denominator?
When you were evaluating the limit, you likely “cancelled” a factor of \((x-4)\) from the numerator and denominator. Why might you have known, before factoring anything, that \((x-4)\) would be a factor shared in the numerator and denominator?
