Show that the limit $\lim_{n\to\infty}\frac{x}{x+n}=0$ for $x\geq 0$

Question

This proof is my first proof for convengence of series of function. I have not seen any example that show how to proof in the book. I did it base on the proof of sequence of number Please check my proof

For $\epsilon >0$ we must find $N$ that $N\leq n$

$$ \frac{x}{x+n}<\frac{x}{n}$$

$$\frac{x}{n}<\epsilon $$

$$\frac{x}{\epsilon }<n$$

Choose $N\geq \frac{x}{\epsilon }$

Then $\frac{x}{x+n}<\frac{x}{n}\leq N< n$

Answer 1

You want to find, given $\varepsilon>0$, some $N$ such that, for $n>N$, $$ \left|\frac{x}{x+n}-0\right|<\varepsilon $$ As $x\ge0$, the inequality can be written $$ x<\epsilon x+\epsilon n $$ so $$ n\ge\frac{x(1-\varepsilon)}{\varepsilon} $$ Take the least positive integer $N$ such that $$ N>\frac{x(1-\varepsilon)}{\varepsilon} $$ and you're done.

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Your method of observing that $$ \frac{x}{x+n}\le \frac{x}{n} $$ is as good, but you should phrase it better.