# Is my proof of $\lim_{x\to \infty}\frac 1x = 0$ correct?

I tried to prove $$\lim_{x\to \infty}\frac 1x = 0$$ I started as thus $$\lim_{x\to \infty}\frac 1x=\lim_{x\to \infty}\frac x{x^2}$$ Applying L'Hospital's Rule $$\lim_{x\to \infty}\frac 1x=\lim_{x\to \infty}\frac x{x^2}=\lim_{x\to \infty}\frac 1{2x}=\frac12\lim_{x\to \infty}\frac 1x$$ Thus, $$\frac12\lim_{x\to \infty}\frac 1x=\lim_{x\to \infty}\frac 1x$$ which therefore implies $$\lim_{x\to \infty}\frac 1x = 0$$ QED.

• For a ninth grader it is sufficient to assume that $\lim_{x\to\infty} \dfrac{1}{x}=0$ without proof. And I wonder what kind of syllabus teaches calculus in ninth grade. – Paramanand Singh Feb 10 '17 at 15:20
• I'm pretty sure you're supposed to work directly from the definition here. First, your proof doesn't prove that the limit exists, but only that if it exists, then it must be $0$. Second, if you need L'Hospital for something as simple as $\lim 1/x$, when how would you prove L'Hospital itself in the first place? – Henning Makholm Feb 10 '17 at 15:20
• @user35508 You can use a thing called the epsilon-delta definition of a limit to prove rigorously. – Simply Beautiful Art Feb 10 '17 at 15:34
• BTW what is the harm in applying L'Hospital's Rule directly on $1/x$. It gives the answer $0$ directly. Why do you multiply by $x$ in numerator and denominator. – Paramanand Singh Feb 10 '17 at 15:50
• If denominator tends to $\infty$ you can apply L'Hospital's Rule. – Paramanand Singh Feb 10 '17 at 15:54

I too tried the same thing:

$$\lim_{x\to\infty}x=\lim_{x\to\infty}\frac{x^2}x\stackrel{L'H}=2\lim_{x\to\infty}x$$

Thus,

$$\lim_{x\to\infty}x=2\lim_{x\to\infty}x$$

And as you have said,

$$\lim_{x\to\infty}x=0$$

QED (?)

• Concluding $y = 0$ from $y = 2y$ only works if $y$ is a real number, which you haven't shown for $\lim_{x\to\infty} x$. (In fact, it isn't.) – Jon Feb 10 '17 at 15:45
• :O And you found the error! (for those who haven't yet seen) – Simply Beautiful Art Feb 10 '17 at 15:46
• Ah, the ol' devil's advocate routine. ;) – Jon Feb 10 '17 at 15:47
• Good example, but this answer should explain the fallacy. The fallacy is that there are unspoken assumptions of L'H which in this case is unmet. – Paul Draper Feb 10 '17 at 18:40
• @PaulDraper see the comments below the second answer :-) – Simply Beautiful Art Feb 10 '17 at 18:50

This is incorrect, as you can only use L'Hospital's Rule when you know the limit of the derivative ratio exists.

• No, L'Hospital's rule was correctly applied here... – Simply Beautiful Art Feb 10 '17 at 15:19
• No, it wasn't. The original rule assumes the limit of the derivative ratio exists, but this is basically what's requested to prove in the original problem. – Theorem Feb 10 '17 at 15:20
• Nope. Showing that $\lim_{x\to\infty}x=\infty$ and $\lim_{x\to\infty}x^2=\infty$ is enough to satisfy the use of L'H. – Simply Beautiful Art Feb 10 '17 at 15:22
• @SimplyBeautifulArt: L'Hospitals rule says (in the appropriate variant) that if $f(x)\to\infty$ and $g(x)\to\infty$ and the limit of $f'(x)/g'(x)$ exists, then the limit of $f(x)/g(x)$ also exists and equals the limit of $f'(x)/g'(x)$. You're ignoring the third premise. – Henning Makholm Feb 10 '17 at 15:33
• @SimplyBeautifulArt: Before L'Hospital tells you anything useful, you need to know already that $f'(x)/g'(x)$ has a limit at all. For example, you can't use L'Hospital on $\frac{x+\sin x}{x}$ to conclude that $$\lim_{x\to\infty}\frac{x+\sin x}{x} = \lim_{x\to\infty}\frac{1+\cos x}{1}$$ in which the LHS is clearly (from first principles!) $1$ whereas the RHS does not converge at all. – Henning Makholm Feb 10 '17 at 15:40

What you have (very cleverly!) shown is that if the limit $\lim_{x\to\infty}{1\over x}$ exists, then, by L'Hopital, it can only equal $0$. Simply Beautiful Art's answer establishes the same result for $\lim_{x\to\infty}x$. The difference is, in your case the limit actually does exist, while in SBA's case it doesn't. That was SBA's implicit message: You haven't proven the limit is $0$, you've only proven a conditional statement; it remains to show that the limit exists.

One possible way to show that the limit exists without explicitly computing it would be to invoke (or prove) a theorem saying that a monotonically decreasing function that's bounded below necessarily has a limit as $x$ tends to infinity.

In essence, you've done the second step of a two-step process. There are other MSE questions where assuming the limit exists allows you to compute it; when I have more time I'll try to provide some links. This is the first time I can think of, though, where I've seen L'Hopital's rule used as part of the derivation.

Not an answer - essentially a comment and too long for a comment that I don't want lost in the flurry of existing comments.

Many students try L'Hopital unthinkingly when faced with the limit of an indeterminate form like $0/0$. Often the application is incorrect. Even when it works it's often not the easiest method, and it's rarely the most illuminating. You learn much more thinking about simple order of magnitude inequalities or the first few terms of Taylor series expansions.

There are many answers on this site that illustrate that. Here are some; other answerers should feel free to edit this answer to link to more.

lHopitals $\displaystyle \lim_{x\rightarrow \infty} \; (\ln x)^{3 x}$?

Finding the limit of a function with a trigonometric exponent

Computing $\lim_{x\to0} \frac 8 {x^8} \left[ 1 - \cos\frac{x^2} 2 - \cos\frac{x^2}4 + \cos\frac{x^2}2\cos\frac{x^2}4 \right]$ without using L'Hospital

Find the limit $\lim_{x\to0}\frac{\arcsin x -x}{x^2}$ The horizontal lines in the picture are $y = \pm \dfrac 12$. As you can see, after $P = 3$ on the $x$ axis, the values of $f(x)$ are contained on the interval $\left(-\dfrac 12, \dfrac 12 \right)$ on the $y$ axis. In informal terms, the rigorous definition of $\lim_{x \to \infty} \dfrac 1x = 0$ is simply the assertion that that you can do exactly what I did above for any horizontal lines $y = \pm \epsilon$, no matter what (positive) $\epsilon$ you pick. That is, for any positive number $\epsilon$, you can always find some point $P$ somewhere on the $x$ axis such that for every $x$ larger than $P$, its $f(x)$ value is on the interval $(-\epsilon, \epsilon)$.