Prove $\lim_{x\to\infty} \frac{1}{x} = 1$ is false

Question

$$\lim_{x\to\infty} \frac{1}{x} = 1$$

Given $\epsilon > 0$ $$\left|\frac{1}{x} - 1\right| < \epsilon.$$ Rewrite it as $$-\epsilon < \frac{1}{x} - 1 < \epsilon$$ $$-\epsilon + 1< \frac{1}{x} < \epsilon + 1$$ If epsilon is very small, then on both sides we are getting value close to $1$, but the function gets closer to zero, hence both sides false. If $\epsilon$ is big, then on the right side we are getting big positive value, but with $ n \in (0,1)$ the function gets bigger too. Hence right side fails. Is this a sound proof? And if yes, how would I rewrite it with math symbols?

Answer 1

I would accept it, because I know what you mean by very small. However, in this case it's best to make precise what you mean. If we use 1/2, and let $x>2$, then $1/x<1/2$. So we can't have$1/x\to1$.

Honestly, the right hand side doesn't matter in this case. We only need to break one of the inequalities to show convergence doesn't hold. But in any case it is always true for $x\geq1$ that $1/x<1+\epsilon$, so the right inequality does hold.

Answer 2

Given $\epsilon > 0$ assume wlog $x>1$ and $\epsilon<1$ then

$$\left|\frac{1}{x} - 1\right| < \epsilon \iff1-\frac1x < \epsilon \iff \frac1x>1-\epsilon \iff x<\frac1{1-\epsilon }$$

then the inequality fails for any $x\ge M=\frac1{1-\epsilon }$.

Answer 3

Your argument that $\frac 1x$ goes to $0$ needs to be proven and that basically is what is being asked to be proven; prove $\frac 1x$ doesn't go to $1$.

ANd if you do prove that $\lim_{x\to \infty} \frac 1x = 0$ (which it does-- see addenda) that's not enough because although the limit notation $\lim_{x\to \infty}f(x) = L$ looks like an equality, it actually means for every $\epsilon > 0$ there is an $N$ so that $x > N \implies |f(x) - L| < \epsilon$ and we don't know that there can't be two so $L$s. (ALthough we can and do prove that very early on-- see addenda).

Here's a hint: $|\frac 1x - 1| =|1-\frac 1x|= |\frac {x-1}x|$.

so if $|\frac 1x - 1|<\epsilon$ then $-\epsilon < \frac {x-1}x < \epsilon$. Now as $x\to \infty$ we can assume $x > 1$ so $-\epsilon x < 0 < x-1 < x\epsilon$

$x-x\epsilon=x(1-\epsilon) < 1$

If we choose an $\epsilon$ so that $0<\epsilon < 1$ we have $x < \frac 1{1-\epsilon}$.

Well, that puts an upper limit on $x$ which contradicts that $x \to \infty$ so that's impossible.

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Addenda:

Claim: $\lim_{x\to \infty} \frac 1x = 0$.

Pf: For any $\epsilon >0$ Let $N =\frac 1{\epsilon}$ (which is positive). If $x > N$ then $|\frac 1x -0| = \frac 1x < \frac 1N =\epsilon$.

Claim: If $\lim_{x\to \infty} f(x) = L$ and $M \ne L$ then $\lim_{x\to \infty} f(x)= M$ is not true.

Proof: If $L \ne M$ then $|L - M| > 0$. Let $\epsilon = \frac {|L-M|}2$

If $|f(x) - M| < \epsilon$ and $|f(x) - L| < \epsilon$ then

$|L - M| = |(L - f(x)) + (f(x) - M)| \le |L-f(x)| + |f(x)-M| < \epsilon + \epsilon = |L-M|$

So $|L-M| < |L-M|$ which is impossible. So there are no $N$ or $N'$ so that if $x >N$ and $x > N'$ (i.e. $x > \max(N,N')$ then $|f(x)-L| < \epsilon$ and $|f(x) -M| < \epsilon$ as that is impossible.

......

So if you didn't want to prove it as I did in the body of this post, you can insteand prove limits, when they exist, are unique. And that $\lim_{x\to \infty}\frac 1x =0$ and that $0 \ne 1$ so the claim $\lim_{x\to \infty}\frac 1x = 1$ is false.

Answer 4

To show that the statement is false, find an $\epsilon$ which does not have a corresponding $x^\star$, such that whenever $x \geq x\star$, $| \frac{1}{x} - 1 | < \epsilon$ does not hold. Suppose $L = 1$ and let $\epsilon = \frac{1}{2}$. Consider when $x^\star \geq 1$, then \begin{align*} | \frac{1}{x} -1 | \geq \frac{1}{2} \end{align*}

whenever $x \geq 2$. Consider when $x^\star <1$, then \begin{align*} |\frac{1}{x} -1 | \geq \frac{1}{2} \end{align*} whenever $x \geq 2$.

Hence, there does not exist an $x^\star$ for $\epsilon = \frac{1}{2}$. Therefore, it must certainly be false that the limit is $1$.

Answer 5

Just as a alternative proof method, consider the improper integral $$I=\lim_{x\to\infty}\displaystyle\int_{1}^{x} \frac{1}{t^2}\,dt$$

We know that since $t^2\geq 0$ for all $t\in\mathbb{R}$, and in this case, $t\geq 1>0$, so we have that $1\geq \frac{1}{t^2}>0$, which implies that the function in the integrand is strictly positive on the interval $[1,\infty)$, so the integral should be strictly positive as well, that is, $I>0$. After computing, we see that $$I=\lim_{x\to\infty} -\frac{1}{t}\bigg|_{t=1}^{t=x} = \lim_{x\to\infty}-\frac{1}{x}+1=\lim_{x\to\infty}-(\frac{1}{x}-1)=-(1-1)=0\not>0$$

So, the assumption that $\lim_{x\to\infty} \frac{1}{x}=1$ is false.