can I say that $\lim(x_n-y_n) = x - y$ if $x=\lim(x_n)$, $y=\lim(y_n)$?

I have this question in my HW: true or false,

If $x_n$ is any increasing sequence of negative real numbers and $y_n$ is a cauchy sequence of real numbers, then the sequence $x_n-y_n$ converges.

My guess it's true. for example $x_n=\{-1/n\}$ which is an example of increasing sequence of real number. In this case $x_n$ converges to $x=0$ and since $y_n$ is cauchy so it converges to a real number $y$, but how can I prove this. I know that $\lim(x_n+y_n)=x+y$ can I use it as $\lim(x_n-y_n)=x-y$?

• en.wikipedia.org/wiki/Limit_of_a_function#Properties – Shuri2060 Jul 9 '17 at 13:18
• Typo in the title. That shall be a $-$? Also, if you can see both $\{x_n\}$ and $\{y_n\}$ converge, then you can safely do arithmetics on them. – Li Chun Min Jul 9 '17 at 13:18
• For proofs, perhaps math.wikia.com/wiki/Algebra_of_limits – Shuri2060 Jul 9 '17 at 13:19
• Welcome to math stack exchange – Peter Jul 9 '17 at 13:21
• Btw, have you heard of monotone convergence theorem? That is actually equivalent to the Cauchy criterion that you have cited. – Li Chun Min Jul 9 '17 at 13:25

The sequence $x_n$ converges because it is increasing and bounded from above by $0$. The sequence $y_n$ converges because it is a Cauchy-sequence and in the real numbers this implies that it converges. As already mentioned in the comments, you can take the limits and subtract them to get the limit of the differnce.
You are right that the Cauchy sequence converges (assuming we are working with sequences in $\mathbb{R}$). The increasing sequence is bounded above by 0, so by the Monotone convergence theorem it also converges.
If you know that $\lim_{n\to\infty} (x_n+y_n) = \lim_{n\to\infty} x_n + \lim_{n\to\infty} y_n$ whenever the limits on the right are defined, then you can easily deduce the corresponding fact about subtraction:
$$\lim_{n\to\infty} (x_n-y_n) = \lim_{n\to\infty} x_n + \lim_{n\to\infty} (-y_n) = \lim_{n\to\infty} x_n - \lim_{n\to\infty} y_n.$$
You do have to believe that if $(y_n)$ converges to $y$ then $(-y_n)$ converges to $-y$, but this is not so hard to prove. It follows that $(x_n-y_n)$ converges, because $x_n$ and $y_n$ do.