If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## Calculus 1

### Unit 1: Lesson 4

Formal definition of limits (epsilon-delta)

# Formal definition of limits Part 4: using the definition

Watch the definition of a limit in action. Created by Sal Khan.

## Want to join the conversation?

• the video was really awesome, thank you so much

just a little question:

is this "ε = 2∂" true for any kind of function or just this example??
• It might have been easier to understand and more universal to have explicitly stated that the relationship between epsilon and delta can be determined using f(x). The function reveals how epsilon and delta are related, just as any other y relates to x as defined by the function. In the example function, y=2x, except at x=5. Substitute epsilon for y and delta for x and you have epsilon=2(delta).
If the function is different than y=2x, the function can still be used to connect epsilon and delta so the substitution can be made in the proof.
• At -, Sal says that x is not equal to 5 as if the expression |x-c| < ∂ exactly implies that x is within range of c but not equal to c. For the latter proposition, shouldn't there also be a 0 < at left-hand side of the inequality ? ( 0 < |x-c| < ∂ ). Because we are trying to approach c, not to get there, given the definition of limit.
• the inequality shown @ will include c. The inequality should read:
0<|x-c|<delta. When you remove the absolute value braces to evaluate
this expression, you will create 2 inequality expressions: one describing
the range x<c, the other the range x<c. The value x=c is excluded.
• I'm so lost from this video
• where did the 5 come from in |x-5| ?
• In the last video, Sal wrote down the equation |x - c| < ∂, which means x is within ∂ of c. |x - 5| < ∂ is the same thing, but we are defining c to be 5.
• What would it look like if the limit you were trying to prove was actually wrong? For example, if you tried to prove that for f(x) = x^2, that as x => 3 that L = 10.
• To disprove a limit, we can show that there is some ∈>0 such that there is no δ>0 such that for all c such that |x-c|<δ, |f(x)-L|<∈.

Let's say ∈<1 (because 3^2=9 and |9-10|=1).
We can always pick c=3 so that |x-c|<δ (because |x-c|=0 and 0<δ), but |f(c)-L|>∈ (because |f(c)-L|=|9-10|=1 and 1>∈).
This means that if ∈<1, there is no δ>0 such that for all c such that |x-c|<δ, |f(x)-L|<∈.

Thus, we have disproved that the limit of x^2 as x approaches 3 is 10.

I hope this helps you disprove limits with the epsilon-delta definition!
• What if it's a function that grows in a non linear way, like in an exponential function?
For exemple, in the function f(x) = 2^x. Given E > 0, would the delta in the left side of a value c be different from the delta in the right side of c?
• but the proof still stands ,that's where the one sided limit proof comes from i think?
• What does he mean when he writes, '|x - c| < delta'?
• This is a formal way of writing that the difference between x and c is less than some extremely small number, delta.
• Why should we prove that for all epsilon if we have a delta then the limit at that point (at which we have to prove the limit) is going to be equal to L(Here L =limf(x) x->a). We can just take the casewhen delta->0 and see whether the epsilon->0.
If epsilon->0 then we can say that within the range of a+delta and a-delta for every x(such that x not equal to a) the value of f(x) approaches L or is approximately equal to L. Isn't that what a limit is?
• |`f(x)` - `L`| < 2𝛿 is almost the same as the inequality |`f(x)` - `L`| < ϵ except the right sides are different. And we know that these two inequalities are equivalent.