# limit as $x$ approaches infinity of $\frac{1}{x}$

How can I show $\lim_{x \to \infty} \frac{1}{x} = 0$ using epsilon delta proof.

Its pretty obvious that the limit is zero, but I am still new at epsilon proofs.

What you are trying to prove is that "if $x$ is large enough, then $\frac{1}{x}$ gets close enough to $0$." Approaching infinity represents the concept of allowing $x$ to be arbitrarily large.

Here is how to state the proof, with some comments along the way. Let $\varepsilon>0$ be some arbitrary real number (this will represent "error" or "distance from 0"). Let $N=\frac{1}{\varepsilon}$ (this represents how close to infinity $x$ must be). Whenever $x>N$, then $\frac{1}{x}<\frac{1}{N}$, and since $\frac{1}{N}=\varepsilon$ we have $\left|\frac{1}{x}-0\right|<\varepsilon$ (that is, the distance between $\frac{1}{x}$ and $0$ is less than the error $\varepsilon$). Therefore, $\lim_{x\to\infty}\frac{1}{x}=0$.

• How can an x greater than an "arbitrary" large N? Feb 15, 2022 at 4:48

Let $f:\Bbb R\setminus\{0\}\to\Bbb R$ be $f(x)=1/x$. You have conjectured that $$\lim_{x\to\infty}f(x)=0$$ Rigorously, this means that for every $\varepsilon>0$ there exists an $M$ such that $$\left\lvert\frac{1}{x}\right\rvert<\varepsilon$$ whenever $x>M$. Can you prove this?

• shouldn't $M$ be defined? Is $N$ a natural number? Mar 8, 2019 at 23:11

For every $$\epsilon > 0$$, there exists $$N > 0$$ such that if $$x > N$$, then $$|\frac{1}{x}|< \epsilon$$.

Finding N, we see, $|\frac{1}{x}|< \epsilon \Rightarrow x > \frac{1}{\epsilon} \\$

Proof: For $$\epsilon > 0$$ , take $$N = \frac{1}{\epsilon}$$ and $$x> N= \frac{1}{\epsilon}$$

$$\frac{1}{x} < \epsilon \Rightarrow |\frac{1}{x} |< \epsilon \\$$

Therefore, $$\lim_{x\to\infty} \frac{1}{x} = 0$$

Hence, proved.