One of the very first things taught in a 1st year Calculus class is the concept of the ‘Limit’. One (simplistic) way of describing it, is by saying that the Limit of the function \(\boldsymbol{f\left(x\right)}\) is what the function ‘wants to be’ as \(\boldsymbol{x}\) approaches a specific value or approaches infinity.
SYNTAX: If we want to find the value of \(\boldsymbol{f\left(x\right)}\) as \(\boldsymbol{x}\) approaches zero, we write: \(\displaystyle{\lim_{x \to 0} f\left(x\right)}\)
Example #1
\begin{align*}
\text{Let’s start with a trivial example: } \lim_{x \to 0} \frac{x + 6}{x + 3} &= \text{???} \\
\text{Here, we just ‘plug in’ zero for } x \text{ and we are done: } \lim_{x \to 0} \frac{x + 6}{x + 3} &= \frac{0 + 6}{0 + 3} = 2
\end{align*}
As the graph above shows, at \(\boldsymbol{x = 0}\), the function ‘wants to be’ \(\boldsymbol{2}\).
Example #2
\begin{align*}
\text{This example is more interesting: } \lim_{x \to 5} \frac{x^{2}\, -\, 3x\, -\, 10}{x\, -\, 5} &= \text{???} \\
\text{‘Plugging in’ } 5 \text{ for } x \text{ doesn’t work: } \lim_{x \to 5} \frac{x^{2}\, -\, 3x\, -\, 10}{x\, -\, 5} &= \frac{5^{2}\, -\, 3\left(5\right)\, -\, 10}{5\, -\, 5} = \frac{0}{0} \quad \text{Undefined!!!} \\
\text{However, if we factor the numerator: } \lim_{x \to 5} \frac{\left(x + 2\right) \left(x\, -\, 5\right)}{x\, -\, 5} &= \lim_{x \to 5} \left(x + 2\right) \text{ and cancel the common factor} \\
\text{Now we can ‘plug in’ } 5 \text{: } \qquad \quad \lim_{x \to 5} \left(x + 2\right) &= 5 + 2 = 7
\end{align*}
As the graph above shows, \(\displaystyle{y = \frac{x^{2}\, -\, 3x\, -\, 10}{x\, -\, 5}}\) is undefined at \(\boldsymbol{x = 5}\); nevertheless, we say that the limit at \(\boldsymbol{x = 5}\) exists, and the limit is \(\boldsymbol{7}\). Looking at it in a simple way, ‘the function wants to be 7’.
Example #3
\begin{align*}
\text{Perhaps an even more interesting example is: } &\lim_{x \to 6} \frac{\sqrt{x + 10}\, -\, 4}{x\, -\, 6} = \text{???} \\
\text{As in the previous example, ‘plugging in’ the limit value doesn’t work: } &\lim_{x \to 6} \frac{\sqrt{x + 10}\, -\, 4}{x\, -\, 6} = \frac{\sqrt{6 + 10}\, -\, 4}{6\, -\, 6} = \frac{0}{0} \quad \text{Undefined!!!} \\
\text{In this case we use the conjugate of the numerator: } &\lim_{x \to 6} \frac{\left(\sqrt{x + 10}\, -\, 4\right)}{\left(x\, -\, 6\right)} \cdot \frac{\left(\sqrt{x + 10} + 4\right)}{\left(\sqrt{x + 10} + 4\right)} \\
\lim_{x \to 6} \frac{\left(x + 10\, -\, 16\right)}{\left(x\, -\, 6\right) \left(\sqrt{x + 10} + 4\right)} = &\lim_{x \to 6} \frac{x\, -\, 6}{\left(x\, -\, 6\right) \left(\sqrt{x + 10} + 4\right)} \\
\text{Cancel the common factor: } &\lim_{x \to 6} \frac{1}{\sqrt{x + 10} + 4} \\
\text{Now we can ‘plug in’ } 6 { : } \; &\lim_{x \to 6} \frac{1}{\sqrt{6 + 10} + 4} = \frac{1}{4 + 4} = \frac{1}{8}
\end{align*}
As the graph above shows, \(\displaystyle{y = \frac{\sqrt{x + 10}\, -\, 4}{x\, -\, 6}}\) is undefined at \(\boldsymbol{x = 6}\); nevertheless, we say that the limit at \(\boldsymbol{x = 6}\) exists, and the limit is \(\boldsymbol{\frac{1}{8}}\).
Example #4
The Limit doesn’t have to be as \(\boldsymbol{x}\) approaches a specific value. In this next example we want to find the value of a function as \(\boldsymbol{x}\) increases without bounds:
\begin{align*}
\lim_{x \to \infty} \frac{5x^{2}\, -\, 3x + 1}{3x^{2}\, -\, 5} &= \frac{\infty}{\infty} \quad \text{What does this mean???} \\
\text{There’s a trick we can use: } \lim_{x \to \infty} \frac{\left(5x^{2}\, -\, 3x + 1\right)}{\left(3x^{2}\, -\, 5\right)} \cdot \frac{\left(\frac{1}{x^{2}}\right)}{\left(\frac{1}{x^{2}}\right)} &= \lim_{x \to \infty} \frac{5\, -\, \frac{3}{x} + \frac{1}{x^{2}}}{3\, -\, \frac{5}{x^{2}}} \\
\text{Now, the fractions with } \mathbf{x} \text{ in the denominator go to zero: } \lim_{x \to \infty} \frac{5\, -\, \frac{3}{x} + \frac{1}{x^{2}}}{3\, -\, \frac{5}{x^{2}}} &= \frac{5\, -\, 0 + 0}{3\, -\, 0} = \frac{5}{3}
\end{align*}
What the above graph shows is that as \(\boldsymbol{x}\) grows without bounds, the graph of \(\displaystyle{y = \frac{5x^{2}\, -\, 3x + 1}{3x^{2}\, -\, 5}}\) approaches \(\displaystyle{\frac{5}{3}}\).
L’Hôpital’s Rule
Well, what about L’Hôpital’s Rule? I’m not going to write about Guillaume L’Hôpital. I will just say that there’s a theorem that bears his name, which states that, under certain conditions, the Limit of the quotient \(\mathbf{\frac{f\left(x\right)}{g\left(x\right)}}\) is determined by the limit of the quotient of the derivatives.
The statement above can be written symbolically as:
\[\lim_{x \to c} \frac{f\left(x\right)}{g\left(x\right)} = \lim_{x \to c} \frac{f’\left(x\right)}{g’\left(x\right)}\]
It is important to realize that this is not about applying the Quotient Rule, but rather, taking the Derivative of the numerator and denominator separately.
Example #2 (again)
\begin{align*}
\text{Applying L’Hôpital’s Rule: } \lim_{x \to 5} \frac{\frac{d}{dx} \left(x^{2}\, -\, 3x\, -\, 10\right)}{\frac{d}{dx} \left(x\, -\, 5\right)} &= \lim_{x \to 5} \frac{2x\, -\, 3}{1} \\
\text{Now we can ‘plug in’ } \boldsymbol{5} \text{: } \lim_{x \to 5} \frac{2x\, -\, 3}{1} = \frac{2\left(5\right)\, -\, 3}{1} &= 10\, -\, 3 = 7
\end{align*}
Example #3 (again)
\begin{align*}
\text{Applying L’Hôpital’s Rule: } \lim_{x \to 6} \frac{\frac{d}{dx}\left(\sqrt{x + 10}\, -\, 4\right)}{\frac{d}{dx}\left(x\, -\, 6\right)} &= \lim_{x \to 6} \frac{\frac{1}{2} \left(x + 10\right)^{-\frac{1}{2}}}{1} \\
\text{Re-writing the derivative: } \qquad \lim_{x \to 6} \frac{\frac{1}{2} \left(x + 10\right)^{-\frac{1}{2}}}{1} &= \lim_{x \to 6} \frac{1}{2 \sqrt{x + 10}} \\
\text{Now we can ‘plug in’ } \boldsymbol{6} \text{: } \qquad \quad \lim_{x \to 6} \frac{1}{2 \sqrt{x + 10}} &= \frac{1}{2 \sqrt{6 + 10}} = \frac{1}{2 \left(4\right)} = \frac{1}{8}
\end{align*}
Example #4 (again)
\begin{align*}
\text{Applying L’Hôpital’s Rule: } \lim_{x \to \infty} \frac{\frac{d}{dx}\left(5x^{2}\, -\, 3x + 1\right)}{\frac{d}{dx}\left(3x^{2}\, -\, 5\right)} &= \lim_{x \to \infty} \frac{10x\, -\, 3}{6x} = \frac{\infty}{\infty} \\
\text{Applying L’Hôpital’s Rule a 2nd time: } \quad \quad \lim_{x \to \infty} \frac{\frac{d}{dx}\left(10x\, -\, 3\right)}{\frac{d}{dx}\left(6x\right)} &= \frac{10}{6} = \frac{5}{3}
\end{align*}
Example #1 (again)
It may be a good time to mention that L’Hôpital’s Rule is applicable only when the Limit is ‘indeterminate’ and of the type: \(\displaystyle{\lim_{x \to c} \frac{f\left(x\right)}{g\left(x\right)} = \frac{0}{0}}\) or \(\displaystyle{\lim_{x \to c} \frac{f\left(x\right)}{g\left(x\right)} = \frac{\infty}{\infty}}\).
\begin{align*}
\text{Recall this trivial example: } \lim_{x \to 0} \frac{x + 6}{x + 3} &= \frac{0 + 6}{0 + 3} = 2 \\
\text{If we insisted in applying L’Hôpital’s Rule, we get: } \lim_{x \to 0} \frac{\frac{d}{dx} \left(x + 6\right)}{\frac{d}{dx} \left(x + 3\right)} &= \lim_{x \to 0} \frac{1}{1} = 1 \quad \text{<=== WRONG!!!}
\end{align*}
Example #5
What to do in those cases when there is not a straight way of finding the Limit and is not in one of the ‘good’ indeterminate types? Let’s see…
\begin{align*}
\lim_{x \to \infty} x \cdot \sin\left(\frac{1}{x}\right) &= \infty \cdot 0 \quad \text{What to do, what to do….?} \\
\text{Convert it into the type } 0/0 \text{ or } \infty / \infty \text{: } \lim_{x \to \infty} x \cdot \sin\left(\frac{1}{x}\right) &= \lim_{x \to \infty} \frac{\sin\left(\frac{1}{x}\right)}{\frac{1}{x}} = \frac{0}{0} \quad \text{Correct type!} \\
\text{Now we can apply L’Hôpital’s Rule: } \lim_{x \to \infty} \frac{\frac{d}{dx}\left(\sin\left(\frac{1}{x}\right)\right)}{\frac{d}{dx}\left(\frac{1}{x}\right)} &= \lim_{x \to \infty} \frac{\left(-\frac{1}{x^{2}}\right) \cos\left(\frac{1}{x}\right)}{-\frac{1}{x^{2}}} = \lim_{x \to \infty} \cos\left(\frac{1}{x}\right) = \cos 0 = 1
\end{align*}
What the above graph shows is that as \(\boldsymbol{x}\) grows without bounds, the graph of \(\mathbf{y = x \cdot \sin\left(\frac{1}{x}\right)}\) approaches \(\boldsymbol{1}\).
Quiz #7
I must admit that I got really excited when I saw this exercise in a Calculus textbook. It took me a while to come up with the answer. In retrospect, it is not very complicated, it only looks like it is.
\begin{align*}
\text{What is this limit? } \quad \lim_{x \to \infty} \frac{\int_{1}^{x} \ln\left(e^{4t\, -\, 1}\right)\: dt}{x}
\end{align*}
Graphs made using Maple 2023. Copyright Maplesoft.
If you find any errors in this post, or have some thoughts about it, please, email me at [email protected] .