Sorry if the question strikes as you as dumb, but the last time I had math in school was 15 years ago and now I'm taking a course, requiring me to do some math.
So in the course material, the teacher solved this one $\lim \limits_{n \to \infty}$ $\sqrt{n+1}$- $\sqrt{n}$ by multiply with a conjugate and she got the result of 0.
I tried solving it, without looking at here solution first and what I did was using the theorems to solve it as follow:
$\lim \limits_{n \to \infty}$$\sqrt{n+1}$ - $\sqrt{n}$ =
$\lim \limits_{n \to \infty}$ $\sqrt{n+1}$ -$\lim \limits_{n \to \infty}$ $\sqrt{n}$ =
$\lim \limits_{n \to \infty}$ $\sqrt{n}$ + $\lim \limits_{n \to \infty}$ $\sqrt{1}$ - $\lim \limits_{n \to \infty}$ $\sqrt{n}$ =
$\lim \limits_{n \to \infty}$ $\sqrt{n}$ + 1 - $\lim \limits_{n \to \infty}$ $\sqrt{n}$ =
1 #
Why is that? and why must I multiply by conjugate? and when should one multiply by conjugate, any specific rules or tips?
I'd be grateful, if you could make your answer as detailed and simple as possible, since as I mentioned before, it's been a while since I had anything to do with math.
Answer
Let's take a step by step look at what you did.
1. $\lim_n (\sqrt{n+1}-\sqrt{n}) = \lim_n \sqrt{n+1}-\lim_n \sqrt n$
This is wrong, since the right hand side is in the indeterminate form $\infty - \infty$. It can be any real number and it can be $\pm\infty$. Thus, there is no sensible way to define it, just like there is no sensible way to define $\frac 00$.
To demonstrate more vividly why this is wrong, let $a$ be any real number. Then $$ a = \lim_n a = \lim_n (n+a-n) = \lim_n (n+a) - \lim_n n = \infty - \infty.$$ If $b$ is another real number, by the same reasoning as above, we get $b = \infty - \infty$. Thus, $a = \infty - \infty = b$, so any two real numbers are equal.
Whoops, we just broke math.
2. $\lim_n \sqrt{n+1} = \lim_n \sqrt n + \lim_n \sqrt 1.$
Ah, this one is not as wrong since contrary to the case $\infty - \infty$, one can actually make sense of $\infty + a = \infty$, for any real number $a$. However $\infty$ is still not a number and you can't do arithmetic with it as usual. For example, $\infty + a = \infty = \infty + b$, for any real $a$ and $b$. Notice that you can't cancel $\infty$ on both sides like you would with some real number $x$ to conclude $a = b$. That would break math once again.
So, if one is careful and knows what they are doing, $\infty + 1 = \infty$ is a fair game. However, in the same line you have $-\lim_n \sqrt n = -\infty$ and you remember what we said about subtracting infinity from infinity: it breaks math.
Thus, your line is still wrong.
Unfortunately, it's likely that it's even more wrong. I'm afraid that you wrote that thinking that $\sqrt{n+1} = \sqrt n + \sqrt 1$. No. Please, no. Nicolas FRANCOIS and I are both having nightmares tonight. Just plug in $n = 8$ and calculate both sides. Doesn't work.
3. $\lim_n \sqrt n + 1 - \lim_n \sqrt n = 1$.
Looks correct, but as I said before, subtracting infinity from infinity breaks math. It would have been correct if it said $\lim_n (\sqrt n -\sqrt n) + 1 = 1$. But it doesn't say that.
TL;DR Please use the rule $\lim_n (a_n+b_n) = \lim_n a_n + \lim_n b_n$ only if it doesn't produce $\infty - \infty$. If $\lim_n a_n = \infty$ and $\lim_n b_n$ is $\infty$ or finite, then $\lim_n (a_n+b_n) = \infty$. If $\lim_n a_n = \infty$ and $\lim_n b_n = -\infty$, then $\lim_n (a_n+b_n)$ can be anything under the heavens.
So, hopefully I've stressed enough that $\infty - \infty$ is a no no. So, how to deal with it?
Observe that $a-b = \frac{(a-b)(a+b)}{(a+b)} = \frac{a^2-b^2}{a+b}$ and so $$\lim_n(\sqrt{n+1}-\sqrt n) = \lim_n\frac{(n+1)-n}{\sqrt{n+1}+\sqrt n} = \lim_n\frac 1{\sqrt{n+1}+\sqrt n} = \left[ \frac 1{\infty + \infty} = \frac 1\infty\right] = 0.$$
This is actually a great strategy whenever you have $\infty - \infty$.
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