Question

Can you find the limit of the following function as x approaches +inf. Can you find the limit of the following function as x approaches +inf. @Mathematics

Answer 1

First of all, does this limit exist and if so what is it?

\[ \lim_{x \rightarrow \infty} (\sqrt{x+1} - \sqrt{x} ) \]

Answer 2

yes it does. According to my wolfram, it is 0

Answer 3

I made a draw and the function actually it goes to zero.

Answer 4

You should see how that's true for yourself. So in any case this suggests that this limit might exist.

But to be more precise, we need to write it in terms of functions we can analyze. Use this identity:

cos a - cos b = - 2 sin((a+b)/2) . sin((a-b)/2)

Answer 5

Hence \( f(x) = \cos\sqrt{x+1} - \cos\sqrt{x} = ...\) what?

Answer 6

yeah I already tried that, howerer I end up with the weirdest thing ever!

Answer 7

Well, your hypothesis is that lim f(x) = 0 as x --> infty

Answer 8

yeah And the answer in the book is zero as well

Answer 9

If you write it out that way then
\[ | \cos\sqrt{x+1} - \cos\sqrt{x} | = 2 | \sin(stuff) \sin((\sqrt{x+1}-\sqrt{x})/2) | \]
\[ \leq 2 |\sin((\sqrt{x+1}-\sqrt{x})/2)| \]

Answer 10

because | sin(stuff) | =< 1

Answer 11

Now you just need to show that \( \sin((\sqrt{x+1}-\sqrt{x})/2) \rightarrow 0 \)

Answer 12

the big question is how. coz if you apply the limit directly, you end up with-inf + inf that is an impossible opperation, isnt it?

Answer 13

Didn't we just say that lim (sort(x+1) - sqrt(x)) = 0?

Answer 14

And what's the limit as a --> 0 of sin(a)?

Answer 15

Yeah I did say that. But I used a program called mathematica to get the answer. I havent proved that (actually I cant)

Answer 16

Here's the proof. Multiply top and bottom by sqrt(x+1) + sort(x) and we have ...

Answer 17

\[ \sqrt{x+1} - \sqrt{x} = \frac{ \sqrt{x+1}^2 - \sqrt{x}^2}{ \sqrt{x+1} + \sqrt{x}} \]
\[ = \frac{ (x+1) - x }{\sqrt{x+1} + \sqrt{x}} = \frac{1}{\sqrt{x+1} + \sqrt{x}} \]

Answer 18

Now the limit of that last term as x --> infinity is zero,
hence the limit of sqrt(x+1) - sqrt(x) = 0 also.

Answer 19

No you are not. I was confused. Yeah it actually tend to zero.

Answer 20

Therefore \[ \lim_{x \rightarrow \infty} \sin((\sqrt{x+1} - \sqrt{x})/2) = 0 \]

Answer 21

Ok. I've got to go and do my own stuff now.

Answer 22

ok thank you so much James