Prove that if $x_n \rightarrow x$ and $x_n + y_n \rightarrow z$, then $y_n \rightarrow z - x$ I've tried a proof of the above question, but it seems too simple to be right. If anyone has some pointers or tips that would be much appreciated as I am rather new to all this.
We can deduce by $x_n + y_n \rightarrow z$ that $y_n$ must converge. So let $y_n \rightarrow y$.
$$x_n + y_n \rightarrow x +y $$
$$x_n + y_n \rightarrow z $$
$$z= x+y $$
$$y= z-x $$
$$y_n \rightarrow y $$
$$y_n \rightarrow z-x $$
Thanks for your time!
 A: Why do you think of $x_n \to x$ and $x_n + y_n \to z$ means $y_n$ converge?
You must show that.
Since $x_n\to x$ that means for any $\epsilon_1 > 0$ then is an $N_{\epsilon_1}$ so that $n > N_1$ implies $|x_n - x| < \epsilon_1$.
And $x_n \to z$ means for any $\epsilon_2 > 0$ then there is an $N_{\epsilon_2}$ so that $n > N_2$ implies $|x_n + y_n- z| < \epsilon_2$.
So if $n > \max (N_{\epsilon_1},N_{\epsilon_2})$ then $|x_n+y_n-z|+|x_n-x| < \epsilon_2 + \epsilon_1$ so 
And meanwhile $|y_n -(z-x)|=|[(x_n+y_n)-z]-[x_n-x]| \le |(x_n+y_n)-z|+|x_n-x|$.
So for any $\epsilon>0$ let $\epsilon_1 = \epsilon_2=\frac \epsilon 2$ and let $N_{\epsilon_1}$ and $N_{\epsilon_2}$ be as above and $N = \max (N_{\epsilon_1},N_{\epsilon_2})$....
then $n > N\implies |y_n-(z-x)=|[(x_n+y_n)-z]-[x_n-x]| \le |(x_n+y_n)-z|+|x_n-x|<\epsilon_2 + \epsilon_1=\frac \epsilon 2 + \frac \epsilon 2=\epsilon$.
So $y_n \to z-x$
A: If you want a formal proof you should go via the $\varepsilon$-$N$ route for sequence convergence. Note that
$$|y_n-(z-x)|=|[(x_n+y_n)-z]-[x_n-x]|\leq |(x_n+y_n)-z|+|x_n-x|.$$
A: As others have said, you can't assume the existence of $y$, the limit of $y_n$. Instead, if you have some algebra of limits theorem proven, you might know that, given $a_n \to a$ an $b_n \to b$, then $a_n + b_n \to a + b$. You might also know that, if $a_n \to a$ and $\lambda \in \Bbb{R}$, then $\lambda a_n \to \lambda a$. Combining these, we see that
$$(-1)x_n \to -x$$
and so
$$y_n = (x_n + y_n) + (-1)x_n \to z + (-1)x = z - x.$$
