Question 1

At 7:23 , how did the limit got inside the logarithm function? It is getting hard for me to make sense for this step. It is like saying lim (x -> 0) cos(x) = cos (lim x->0 x).
How is that possible?
Can this thing be only applied to logarithm functions or is it generic for other functions also like cos, sin etc?

Accepted Answer

It's NOT a general rule, and I wish Sal spent some time explaining why it works in this _particular case._

– – –

First of all, we're dealing with a _composite function._

𝑓(𝑥) = 1∕ln 𝑥
𝑔(𝑥) = (1 + 𝑥)^(1∕𝑥)
⇒
𝑓(𝑔(𝑥)) = 1∕ln((1 + 𝑥)^(1∕𝑥))

In general terms we are looking for
𝐹 = lim(𝑛 → 0) 𝑓(𝑔(𝑛))

This means that we let 𝑛 approach zero, which makes 𝑔(𝑛) approach some limit 𝐺, which in turn makes 𝑓(𝑔(𝑛)) approach 𝐹.

In other words:
𝐺 = lim(𝑛 → 0) 𝑔(𝑛)
𝐹 = lim(𝑔(𝑛) → 𝐺) 𝑓(𝑔(𝑛)) = [let 𝑥 = 𝑔(𝑛)] = lim(𝑥 → 𝐺) 𝑓(𝑥)

Now, if we use our definitions of 𝑓(𝑥) and 𝑔(𝑥), we get
𝐺 = lim(𝑛 → 0) (1 + 𝑛)^(1∕𝑛) = [by definition] = 𝑒
𝐹 = lim(𝑥 → 𝑒) 1∕ln 𝑥 = [by direct substitution] = 1∕ln 𝑒 = 1

Note that 𝐹 was given to us by direct substitution, which means that in this _particular case_ we have
lim(𝑥 → 𝐺) 𝑓(𝑥) = 𝑓(𝐺) = 𝑓(lim(𝑛 → 0) 𝑔(𝑛))

– – –

EDIT (10/28/21):
The reason this works is because lim 𝑥→0 𝑔(𝑥) = 𝑒 (i.e. the limit exists)
and𝑓(𝑥) is continuous at 𝑥 = 𝑒

According to the theorem for limits of composite functions we then have
lim 𝑥→0 𝑓(𝑔(𝑥)) = 𝑓(lim 𝑥→0 𝑔(𝑥))

Sal explains that theorem here:
https://www.khanacademy.org/math/ap-calculus-ab/ab-limits-new/ab-1-5a/v/limits-of-composite-functions

Question 2

How can e^x be the only function that is the derivative of itself? Doesn't f(x) = 19e^x also satisfy this property?

Accepted Answer

When we say that the exponential function is the only derivative of itself we mean that in solving the differential equation f' = f. It's true that 19f = (19f)' but this isn't simplified; I can still pull the 19 out of the derivative and cancel both sides. You are correct in saying that the general solution is Ae^x where A is a real value; however, the "A" part isn't the main focus - the main focus is the exponential, since that's what varies and the constants don't.

Question 3

Where can I find the proof of limit as n→infinity (1+1/n)^n =e and limit as n→0 (1+n)^(1/n)=e?

Accepted Answer

https://www.khanacademy.org/math/algebra2/exponential-and-logarithmic-functions/e-and-the-natural-logarithm/v/e-as-limit

or

https://mathcs.clarku.edu/~djoyce/ma122/elimit.pdf

The proof of the two formulas are the same:
lim_{n → ∞} (1 + 1/n)^n = lim_{1/n → 0} (1 + 1/n)^(n) = lim_{x → 0} (1 + x)^(1/x).

Question 4

how/why is (1+1/n)^n equal to (1+n)^(1/n)? Is this just a basic law of exponents

Accepted Answer

Think about it like this:

it is completely legal for us to define one variable as some amount of another variable. Therefore, we can say that n=1/u, for example.

Let's say n=1/u

and

(lim n-> inf) e= (1+1/n)^n

Now let's rewrite this in terms of u. The limit will be that u  gets very small and approaches 0, because this will cause the fraction 1/u to become very large. For n=1/u: if n approaches infinity, u must approach 0 for both sides to approach infinity.

(lim u-> 0) (1+u)^(1/u)   (I simplified 1/(1/u) to just u)

This, therefore, is equivalent to the other definition of e, because all we have done is described the variable in a new way without adding in or taking away anything from the original equation, just looking at it differently.

Question 5

At 7:23, is it that this is an application of the principle:

lim(x->a)[ f(g(x)) ] = f( lim(x->a)[g(x)] )

?

Accepted Answer

Yes, with 𝑓(𝑥) = ln 𝑥 and 𝑔(𝑥) = (1 + 1∕𝑥)^𝑥
we get 𝑓(𝑔(𝑥)) = ln(1 + 1∕𝑥)^𝑥

Because the natural log function is continuous, we have
lim[𝑥 → ∞] 𝑓(𝑔(𝑥)) = 𝑓(lim[𝑥 → ∞] 𝑔(𝑥))
= ln(lim[𝑥 → ∞] (1 + 1∕𝑥)^𝑥)

Question 6

Hi - i am interested that sal says that e = (1+n)^1/n  when I graphed y = (1+x)^1/x the graph converges to 1.  What mistake have I made?

Accepted Answer

What you may have missed is lim (n->0) for that definition. You are correct that lim (n->∞) (1+x)^1/x = 1, but lim (n->0) (1+x)^1/x = e.

Question 7

Technically, the function x^0-1 is its own derivative.

Accepted Answer

Any function of the form a·e^x is its own derivative, and these are the _only_ functions that are their own derivatives. The zero function is just the special case where a=0.

Question 8

1. at 5:10, can you rigorously prove Delta x->0 equals n->0?

2. at 7:17, which limit property allows:
lim n->0 ln ((1+n)^1/n) (denominator) 
to become:
ln (lim n->0 (1+n)^1/n)?

Thanks.

Accepted Answer

1. Ooo that's a hard one! I tried a proof using a method called the "epsilon-delta method" (which is the most rigorous method of proving literally anything related to limits) and it does seem to work. Here it is: *WARNING* This proof is pretty long and exhaustive. I'd definitely recommend learning the epsilon-delta method first and then going over this. It's available later on in the course. So, we have $\Delta$x = ln(n+1). We are given the statement that as $\Delta$x tends to zero, n tends to 0 as well. We can prove these two separately. We assume the statement is true. And if we can prove that both $\Delta$x and ln(n+1) tend to the same number, we should be done. Now, let's first prove that $\lim\limits_{\Delta x o 0} \Delta$x = 0. Now, if I let $\epsilon$>0, I will have a $\delta$>0 such that if $\mid \Delta x - 0 \mid < \epsilon$, $\mid \Delta x - 0 \mid < \delta$. We don't have to do much here as this inequality is true when $\delta = \epsilon$. So, we've proven the first limit. Now, the second one is pretty hard. We have $\lim\limits_{n o 0} ln(n+1)=0$. Now again, if we have $\epsilon>0$, we can find a $\delta>0$ such that if $\mid ln(n+1) - 0 \mid < \epsilon$, $\mid n - 0 \mid < \delta$. Now, we start with $\mid ln(n+1)-0 \mid<\epsilon$. We can write this as $\mid ln(n+1) - ln(1)\mid< \epsilon$. This can be expanded to $-\epsilon < [ln(n+1)-ln(1)]<\epsilon$. Now, using log properties, we get $-\epsilon

Calculus 1

Course: Calculus 1 > Unit 2

Proof: The derivative of 𝑒ˣ is 𝑒ˣ

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