Question

Show that \[\sqrt{x+2}\rightarrow2 \] as \[x \rightarrow 2\]

Answer 1

Are you looking for a hardcore proof?

Answer 2

Yeah, from the definition of a limit. Should have absolute values etc.

Answer 3

I could do it if it was, say a factorisable quadratic, but with the sqrt, i'm a bit confused.

Answer 4

where x=0, sqrt of 2 is like 1.41 , where x = 1 sqrt of 3 is between 1.41 nd 2, just before x=2, the sqaure gets closer to 4 and the sqrt gets closer and closer to 2

Answer 5

Alright, give me a sec

Answer 6

but yeah thats ignoring negative values obviously

Answer 7

Maybe even the first couple of steps re working with the sqrt function. I'd kind of like to figure most of it out myself.

Answer 8

Let
\[\epsilon > 0\]
be given. We seek some
\[ \delta > 0 \]
such that
\[ |x - 2| < \delta \text{ implies that } |\sqrt{x+2}-2| < \epsilon \]
Let's restrict delta as follows:
\[ \delta < 1 \]
which means that
\[ |x -2 | < 1 \rightarrow 1 < x < 3 \]
Further, note that
\[|x - 2| = |x+2 -4| = |\sqrt{x+2} -2 | \cdot |\sqrt{x+2} + 2| \]
since
\[ 1 < x < 3, |\sqrt{x+2} + 2| > \sqrt{3}+2 \]
This means that
\[ |x-2| > |\sqrt{x+2}-2| \cdot (\sqrt{3}+2) \]
Therefore, let
\[ \delta = min \left( 1 , \epsilon\cdot (\sqrt{3}+2) \right) \]
So that if
\[ |x-2| < \delta\]
we have
\[|x-2| = |\sqrt{x+2} -2| \cdot |\sqrt{x+2} -2| > |\sqrt{x+2}-2|\cdot (\sqrt{3}+2)\]
\[ \text{ is less than } \epsilon \cdot (\sqrt{3}+2) \text{ which implies that } \]
\[ |\sqrt{x+2} -2| < \epsilon \]

whew. Hopefully I didn't make a typo...

Answer 9

Almost definitely did though ;)

Answer 10

Also oops, I didn't read your above message.... restrict yourself to the first few lines if you can haha

Answer 11

Where did the x-2 come from?

Answer 12

It's the limit as x approaches 2

Answer 13

so |x-2| gets tiny

Answer 14

So x gets within delta of 2, ok, gotcha.

Answer 15

I get delta = min{1, (epsilon/(sqrt3+2))

Answer 16

Yes you're right, I'm sorry. Told ya I probably made a typo somewhere....

Answer 17

Actually, does it matter that you used the lower bound of x. If you use the 3, does it still hold?

Answer 18

The way I wrote it, no, because of the > and < signs. Choosing delta < 1 was arbitrary though. Just for convenience.

Answer 19

since
\[ |x-2| = |\sqrt{x+2}-2| \cdot |\sqrt{x+2}+2| > |\sqrt{x+2}-2| \cdot (\sqrt{3}+2)\]
we have that
\[ |\sqrt{x+2}-2| < \frac{|x-2|}{\sqrt{3}+2} \]
so if |x-2| < delta, where
\[ \delta = \epsilon \cdot (\sqrt{3}+2) \]
then
\[| \sqrt{x+2}-2|< \frac{|x-2|}{\sqrt{3}+2} < \frac{\epsilon \cdot (\sqrt{3}+2)}{\sqrt{3}+2} = \epsilon \]