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## Calculus, all content (2017 edition)

### Course: Calculus, all content (2017 edition) > Unit 1

Lesson 12: Limits from equations (direct substitution)# Undefined limits by direct substitution

AP.CALC:

LIM‑1 (EU)

, LIM‑1.D (LO)

, LIM‑1.D.1 (EK)

Sal gives an example of a limit where direct substitution ends in a quotient with 0 in the denominator and non-0 in the numerator. Such limits are undefined. What about limits where substitution ends in 0/0? Keep going and you'll see!

## Want to join the conversation?

- does undefined limits by direct substitution always has 0 answer?(3 votes)
- When you have to divide something by zero, it's gonna be undefined.(39 votes)

- what's the full function of ln?(7 votes)
- The function ln(x) is called the natural logarithm. It's a logarithm with a base of the mathematical constant
**e**(About 2.718281). This function can also be written as f(x) = log(base**e**) x.

Hope this answers your question :)(24 votes)

- Do limits exist for irrational or imaginary numbers/expressions?(13 votes)
- Absolutely. For irrational expressions, consider for example the function f(x) = x + pi. That is a continuous function for which the limit approaching any value of x will be x + pi (an irrational number).

Complex functions (i.e. involving imaginary numbers) behave just the same in the sense that they can have limits defined, and those limits can be complex numbers. Simple example: The limit of f(x) = ix as x approaches 1 is i.(4 votes)

- What about L'Hopital's rule where you can take the derivative of the numerator and denominator and try to find the limit of that. In which case you would get lim as x approaches 1 of 1/(1/x) which is just x which makes the limit 1(4 votes)
- The issue of L'Hoptials rule is that it really only works for indeterminate forms, such as 0/0 (the other being infinity / infinity).

I'm sure the proof of this is decently involved, and I'd like to check it out now myself. But for the most part you can only use it for indeterminate forms.

See the simple example of x / 1 as x -> 0

If we used L'Hopitals rule we get: 1 / 0... well that's nowhere near correct.

What about x^2 / [x + 3] as x -> 2

If we use L'Hopital's rule we get: 2x / 1 x->2, plugging in with direct substitution we get 4/1

If we evaluated the original limit we get 2^2 / [2+3] or 4/5, this is a much different limit.

The scenario where we have 0/1 or 1/0 is much the same case, since they're not necessarily indeterminate forms. You may, however, bring a formula that is in a different form TO indeterminate form via manipulation, but you must ensure you do that first BEFORE applying L'Hopital's rule.(10 votes)

- Wait, what about the limit of 1/x² when x approaches 0? It tends to infinity, though direct substitution lead to 1/0. What exactly does make this method
**not**work there? The discontinuity?(3 votes)- You could say that it tends to infinity, but infinity is not a number. So you can never say "something = infinity". (Well, I don't know about
**never**, but at least not in the context of limits.)(9 votes)

- Is it possible that the value of the function doesn't exist at that specific point but there is still a limit like what was mentioned in the previous videos?(5 votes)
- Yes! As long as the limit is a real number or an infinite, then there is a limit no matter the existence of the point.(2 votes)

- 0:12Why is lim x->1 (x/ln(x)) equals lim x->1 (x) /lim x->1 (ln(x)) ?(2 votes)
- It's a property of the limits! Check this video: https://www.khanacademy.org/math/differential-calculus/dc-limits/dc-limit-prop/v/limit-properties(4 votes)

- I didn't get why top limit evaluates to 1. Can someone explain clearly?(2 votes)
- Let 𝑓(𝑥) = 𝑥

Thereby,

lim(𝑥 → 1) 𝑥 = lim(𝑥 → 1) 𝑓(𝑥)

Since 𝑓(𝑥) is continuous we can use direct substitution, and we get

lim(𝑥 → 1) 𝑓(𝑥) = 𝑓(1) = 1(3 votes)

- At1:01, how does Sal just know all this stuff about the natural log? How does he know that the natural log isn't defined for all x's, and how does he know that ln(x) IS continuous at x=1?(2 votes)
- Some basic properties of the logarithmic function should be known before learning limits. In particular, the natural log function is continuous at 𝑥 = 1 as the inverse (exponential function) is continuous at 𝑥 = 0. Furthermore, there are domain restrictions to the natural log function as there are range restrictions in its inverse. If you are unsure about these properties, I would suggest watching the KA videos on inverse functions, logarithms, and exponential functions. Comment if you have questions!(3 votes)

- If the limit is 0, does that mean it doesn't exist?

I feel like it does exist but I'm not sure.(2 votes)- Is zero determinate? Yes, so the limit exists.(2 votes)

## Video transcript

- [Voiceover] Let's see if
we can figure out the limit of x over natural log of
x as x approaches one. And like always pause this
video and see if you can figure it out on your own. Well we know from out limit
properties this is going to be the same thing as the limit as x approaches one of x over over the limit, the limit as x approaches one of the natural log of x. Now this top limit, the
one I have in magenta, this is pretty straight forward, if we had the graph of y equals x that would be continuous everywhere it's defined for all real
numbers and it's continuous at all real numbers. So it's continuous to limit
as x approaches one of x. It's just gonna be this
evaluated x equals one. So this is just going to be one. We just put a one in for this x. For the numerator here we
just evaluate to a one. And then the denominator, natural log of x is not
defined for all x's, therefore it isn't continuous everywhere. But it is continuous at x equals one. And since it is continuous
at x equals one, then the limit here is just
gonna be the natural log evaluated at x equals one. So this is just going
to be the natural log the natural log of one. Which of course is zero. E to the zero power is one. So this is all going to be equal to this is going to be equal to we just evaluate it one over one over zero. And now we face a bit of a conundrum. One over zero is not defined. It is was zero over zero,
we wouldn't necessarily be done yet but it's indeterminate form as we will learn in the future
there are tools we can apply when we're trying to find
limits and we evaluate it like this and we get zero over zero. But one over zero. This is undefined which tells us that this limit does not exist. So does not exist. And we are done.