$\lim_{n\to \infty}\frac{1}{n} \sum_{k=1}^n x_k =x\;$ given $\;\lim_{n\to \infty} x_n= x\;?$

I have a question which is giving me a hard time.

I want to show that $$ \lim_{n\to \infty}\frac{1}{n} \sum_{k=1}^n x_k =x$$ given that $\lim_{n\to \infty} x_n= x$.

Or you could slog through a tedious proof:

Choose $\epsilon>0$. Let $N$ be such that $n\geq N$ means $|x_n-x| < \frac{\epsilon}{2}$. Now choose $N'\geq N$ so that $n\geq N'$ means $\frac{1}{n} \sum_{k=1}^N |x_n-x| < \frac{\epsilon}{2}$.

Then, if $n\geq N'$, we have the estimate:

$$|\frac{1}{n} \sum_{k=1}^n (x_n-x)| \leq \frac{1}{n} \sum_{k=1}^n |x_n-x| \leq \frac{1}{n} \sum_{k=1}^N |x_n-x| + \frac{1}{n} \sum_{k=N+1}^n |x_n-x| < \frac{\epsilon}{2}+n\frac{1}{n}\frac{\epsilon}{2}= \epsilon$$

Hence $\lim_{n\to\infty} \frac{1}{n} \sum_{k=1}^n (x_n-x) = 0$ from which the result follows.

Hint: Lookup Cesàro summation.