Confusing Notation Khan Academy

Question

The above image is from Khan Academy. I am confused by the fact that the text says "$x=t$". The top expression has the notation ${x\to t}$ so how can $x=t$ at the same time. $x$ approaching $t$ is not the same thing as $x$ being equal to $t$.

I am also confused by this:

The correct answer ends up being A. But in order to get A, $e$ must equal $t$ if we follow the second expression in the first image I posted. So we can rewrite $f'(e)$ as $f'(t)$. I always see the derivative and its function having the same input, for example $g(x)$ and $g'(x)$, where the input for both the function and its derivative is x. But in the case of this question, we have $f(x)$ and $f'(t)$, and we could never draw these two functions on the same graph because of the different horizontal axes. What am I missing?

Answer 1

This kind of language use has often confused me. It is clearer to say the following:

First Definition of Differentiability:

Suppose we are given a function $g:\mathbb{R} \to \mathbb{R}$ and a number $t \in \mathbb{R}$. We say $g$ is differentiable at the point $t$ if the limit \begin{equation} \lim_{x \to t} \dfrac{g(x) - g(t)}{x-t} \end{equation} exists. In this case, we denote $g'(t)$ to be this limit, and we call $g'(t)$ "the derivative of $g$ at $t$."

"Second" Definition of Differentiability:

Suppose we are given a function $g:\mathbb{R} \to \mathbb{R}$ and a number $t \in \mathbb{R}$. We say $g$ is differentiable at the point $t$ if the limit \begin{equation} \lim_{h \to 0} \dfrac{g(t+h) - g(t)}{h} \end{equation} exists. In this case, we denote $g'(t)$ to be this limit, and we call $g'(t)$ "the derivative of $g$ at $t$."

As mentioned in the image from Khan Academy, both these limit expressions are equivalent, which means these "two" definitions are really just different ways of saying the same thing. This is why I put the word 'second' in quotation marks above.

Now, pay close attention to my choice of words in the definition. To define the concept of "differentiability of $g$ at $t$", we only need two pieces of information: the function $g$ and a number $t$. In the definition, there is no significance to the symbols "$x$" and "$h$"; these are just "dummy variables" for the limit. I could have just as well said $g$ is differentiable at $t$ if the limit \begin{equation} \lim_{\xi \to t} \dfrac{g(\xi) - g(t)}{\xi -t} \end{equation} exists. Or equivalently, if the limit \begin{equation} \lim_{\eta \to 0} \dfrac{g(t + \eta) - g(t)}{\eta} \end{equation} exists.

Just to make sure the letter "$x$" or any other letters don't cause any confusion, consider the following question and answers:

Question: Define what it means for a function $g$ to be differentiable at the point $x$.

Answer 1: $g$ is differentiable at $x$ if $\lim\limits_{\mu \to x}\dfrac{g(\mu) - g(x)}{\mu - x}$ exists.

Answer 2: $g$ is differentiable at $x$ if $\lim\limits_{@ \to 0}\dfrac{g(x + @) - g(x)}{@}$ exists.

Both these answers are correct.

Now, onto the second question. I hope you understand that the phrase $f(x) = e^x$ is a short way of saying "$f : \mathbb{R} \to \mathbb{R}$ the function, which takes a number $x$ as input and gives the number $e^x$ as output." We can express this as $f(\xi) = e^{\xi}$ or as $f(@) = e^@$, so once again, there is no significance attached to the symbol appearing inside the brackets, because it can be anything you like. You're being asked to write $f'(e)$ as a limit. So, we just apply the ("second") definition: \begin{align} f'(e) &= \lim_{h \to 0}\dfrac{f(e+h) - f(e)}{h} \\ &= \lim_{h \to 0}\dfrac{e^{e+h} - e^e}{h} \end{align}

Conclusion:

The phrase "the derivative of the function $g$ at the point where $x=t$ ..." is just sloppy and misleading, it is more accurate to say something like "the derivative of the function $g$ at the point $t$ ...", which is what I said in the definitions above.

I hope I've provided enough "weird" examples of notation with the use of $\xi$, $\eta$, @ that you understand the difference between a "dummy variable", and the actual point of interest.