Homework 2 Problem 2

I have been struggling with problem number 2 from Homework 2 and was hoping for some hints or suggestions about how to approach this problem....I am stumped.

The problem states:

Suppose that $(x_n)$ and $(y_n) $ are sequences with $x_n \rightarrow L$ and $\left|x_n - y_n\right| < \frac{1}{n} $ for all $n$. Show that $y_n \rightarrow L$.

I have toyed around with adding and subtracting L and then using the triangle inequality to say:

$$\left|x_n - y_n\right| =\left|x_n-L+L- y_n\right| \leq \left|x_n-L\right|+\left|L-y_n\right| = \left|x_n - L\right| + \left|y_n - L\right|$$
Since I know that $x_n \rightarrow L$, then $\left|x_n -L\right| < \epsilon.$ But how can I use any of this to show $$\left|y_n - L\right| < \epsilon?$$

Any tips or hints or pointers in the right direction?

Try toying around with adding and subtracting $x_n$ instead, but from $|y_n - L|$, and choose an $N$ that ends up satisfying $\dfrac{1}{N} < \dfrac{\epsilon}{2}$.

Here's what I told a few folks in the office today about the "thought process". There are generally two kinds of things you should always write down:

1. Things you can control, and
2. things you want to control.

By "control", I mean that you can control its size. In this case, the two things that you can control are
$$|x_n-L|$$
(because $x_n\rightarrow L$) and
$$|x_n - y_n|$$
(since this is bounded by $1/n$ by assumption).
As often happens in limit proofs, there is exactly one thing that you want to control, namely
$$|y_n-L|.$$
Generally, you should start with the thing you want to control and try to write it in terms of the things that you have control of. Thus, rather than starting with $|x_n-y_n|$ (as lots of people have), you should start with $|y_n-L|$. Let's see what happens if we add and subtract an $x_n$:
\begin{align}
|y_n-L| &= |y_n - x_n + x_n - L| \\
&\leq |y_n-x_n| + |x_n-L|.
\end{align}
Since we have control over those last two terms, we can make them as small as we like - smaller, than oh let's say, $\varepsilon/2$.

This "thought process" for limit proofs helps. I think I have gotten too used to NOT starting with what I want to "show" that I am shying away from starting with what I want to "control", if that makes sense. Thanks...

Now that the homework's turned in I figured I'd put this out for everyone's opinion. This is my submission for number two. I think the sub-claim bit was overkill but here it is!

Let $\epsilon >0$ and choose $N \in \mathbb{N}$ such that $N> \frac{1}{\epsilon}$. Recall that $| x_{n} - y_{n} | < \frac{1}{n} $. It follows that $y_{n}$ converges since clearly this inequality would not hold for all $n$ if $y_{n}$ diverged.

Sub-Claim: $| | x_{n}-y_{n} |-0| =| (x_{n}-y_{n})-0|$

\textit{Sub-Proof:}

$$| |x_{n}-y_{n}|-0|= | |x_{n}-y_{n}||$$

$$=|x_{n}-y_{n}|=|(x_{n}-y_{n})-0|.$$

Now, since $ | x_{n}-y_{n} | < \frac{1}{n}$, we can conclude from our sub-proof that,

$$|(x_{n}-y_{n})-0| < \frac{1}{n} < \epsilon$$

for all $n > N$. Therefore, the limit of $(x_{n} - y_{n})$ is zero. From theorem 2.3.3 we know,

$$\text{lim}(x_{n}-y_{n})=\text{lim}(x_{n}) - \text{lim}(y_{n})$$

It follows that

$$0=L-\text{lim}(y_{n})$$

$$\text{lim}(y_{n}) = L.$$