Question

demonstrate:
1) lim x -> 0 + sqrt x = 0
2) lim x → 0 f (x) = L ⇔ lim x → 0 (f (x) - l) = 0

Answer 1

Are we going to use the epsilon delta definition?

Answer 2

The limit: \[
\lim_{x\to a}f(x) = L
\]is defined as: \[
\forall \epsilon\;\exists\delta\quad\forall x:\;0\lt |x-a|\lt\delta\implies |f(x)-L|\lt \epsilon
\]

Answer 3

I need help as not prove either

Answer 4

So you don't need to prove?

Answer 5

The only way I know to 'demonstrate' it is using Epsilon Delta.

Answer 6

Not as demonstrate these problems and are for tomorrow

Answer 7

not as... as not.... what am I supposed to do?

Answer 8

if you could help me I'd appreciate it

Answer 9

How do you want help?

Answer 10

if you can you do the shows to see how you do

Answer 11

The right-hand limit: \[
\lim_{x\to a^+}f(x) = L
\]is defined as: \[
\forall \epsilon\;\exists\delta\quad\forall x:\;0\lt x-a\lt\delta\implies |f(x)-L|\lt \epsilon
\]

In this case \[
\lim_{x\to 0^+}\sqrt{x} = 0
\]is: \[
\forall \epsilon\;\exists\delta\quad\forall x:\;0\lt x-0\lt\delta\implies |\sqrt{x}-0|\lt \epsilon
\]

Answer 12

We simplify a bit: \[
\forall \epsilon\;\exists\delta\quad\forall x:\;0\lt x\lt\delta\implies |\sqrt{x}|\lt \epsilon
\]Since \(x>0\) we can modify the epsilon inequality: \[
\forall \epsilon\;\exists\delta\quad\forall x:\;0\lt x\lt\delta\implies |\sqrt{x}||\sqrt{x}|\lt \epsilon^2
\]\[
\forall \epsilon\;\exists\delta\quad\forall x:\;0\lt x\lt\delta\implies x\lt \epsilon^2
\]

Answer 13

Notice how the inequality holds when \(\delta = \epsilon^2\)

Answer 14

So when they give us an \(\epsilon > 0\) we give them \(\delta =\sqrt{\epsilon}\)

Answer 15

This demonstrates\[
\lim_{x\to 0^+}\sqrt{x} = 0
\]

Answer 16

Now to demonstrate:\[
\lim_{x \to 0} f (x) = L \iff \lim_{x \to 0} (f (x) - l) = 0
\]

Answer 17

\[
\lim_{x \to 0} f (x) = L \iff \lim_{x \to 0} (f (x) - L) = 0
\]

Answer 18

First let's say that \(g(x) = f(x)-L\). \[
\lim_{x \to 0} f (x) = L \iff \lim_{x \to 0} g(x) = 0
\]

Answer 19

\[
\lim_{x \to 0} f (x) = L
\]By definition is: \[

\epsilon>0\;\exists\delta>0\quad\forall x:\;0\lt |x-0|\lt\delta\implies |f(x)-L|\lt \epsilon
\]Which is the same as: \[
\epsilon>0\;\exists\delta>0\quad\forall x:\;0\lt |x-0|\lt\delta\implies |g(x)|\lt \epsilon
\]Which is the same as\[
\epsilon>0\;\exists\delta>0\quad\forall x:\;0\lt |x-0|\lt\delta\implies |g(x)-0|\lt \epsilon
\]Which gets us back to \[
\lim_{x \to 0} g(x) = 0
\]

Answer 20

Thank you thank you