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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.

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    I don't think it is correct because you don't know if the desired limit exists or not. – 2017-02-10
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    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. – 2017-02-10
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    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? – 2017-02-10
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    I would say that the limit is pretty much self-evident (but of course there is some rigorous way to prove it). Also, just as a grammatical side-note, you can't really say "I started as thus".. "Thus" can be understood as a synonym for "therefore". – 2017-02-10
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    (Not that it isn't an ingenious manipulation, mind you ... just that it's a detour all the same). – 2017-02-10
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    How to prove the given question rigorously?? – 2017-02-10
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    @user35508 You can use a thing called the epsilon-delta definition of a limit to prove rigorously. – 2017-02-10
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    Ok thanks... I think I can manage with epsilon delta definition but I still don't understand what is wrong in the above proof... – 2017-02-10
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    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. – 2017-02-10
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    As for L'Hospital rule..the ratio should be indeterminate – 2017-02-10
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    If denominator tends to $\infty$ you can apply L'Hospital's Rule. – 2017-02-10
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    L'Hospital's Rule can be applied either for $0/0$ form or when absolute value of denominator tends to $\infty$. – 2017-02-10
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    You need yo update your knowledge of L'Hospital's Rule. See wiki article for this rule – 2017-02-10
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    It appears you did not check Wikipedia article – 2017-02-10
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    @ParamanandSingh lol, that case... – 2017-02-10
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    @SimplyBeautifulArt: using L'Hospital's Rule for this simple problem is too much of an overkill. I don't understand how the rule became the most widely used method on MSE for limit evaluation. – 2017-02-10
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    @ParamanandSingh and likewise we have complaints over limits without L'H :'-( – 2017-02-10
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    You have proved that **if** the limit exists, then is zero. – 2017-02-10
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    @Martín-BlasPérezPinilla *If the limit exists and* ***is finite***, *then it is* $0$. – 2017-02-10
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    @egreg, saying that the limit exist when is $\infty$ is a (very popular) abuse. Almost all the useful properties of the limits are lost if $\infty$ is accepted. – 2017-02-10

6 Answers 6

35

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 (?)

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    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.) – 2017-02-10
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    :O And you found the error! (for those who haven't yet seen) – 2017-02-10
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    Ah, the ol' devil's advocate routine. ;) – 2017-02-10
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    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. – 2017-02-10
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    @PaulDraper see the comments below the second answer :-) – 2017-02-10
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    @SimplyBeautifulArt, well if you think my attribution of the fallacy to be wrong, choose what you think to the fallacy. Either way, a "the proof is wrong because this similar proof is wrong" isn't as helpful an answer at it could be. – 2017-02-10
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    @PaulDraper the idea is to show in a not so bad manner where the OP made jumps in the proof. – 2017-02-10
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    @SimplyBeautifulArt, which I applaud. Specifically, what is the jump? – 2017-02-10
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    @PaulDraper See the first comment below my answer. – 2017-02-10
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    @SimplyBeautifulArt - when referring to another answer, please use the name of the poster. Which post is "second" depends on the sorting order chosen, and for "votes", can change over time. – 2017-02-10
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    @PaulSinclair Oh, sorry, my bad. – 2017-02-10
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    @PaulDraper See the comments below Theorem's answer. – 2017-02-10
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    No problem - just a pet peeve of mine. I occasionally find threads where someone has made such a reference that no longer holds, and it requires an intensive search the find the correct post so that you can follow what is being said. – 2017-02-10
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    Yeah, sorry, just happens without my thinking... – 2017-02-10
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This is incorrect, as you can only use L'Hospital's Rule when you know the limit of the derivative ratio exists.

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    No, L'Hospital's rule was correctly applied here... – 2017-02-10
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    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. – 2017-02-10
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    No, I'm fairly certain that if $\lim_{x\to\infty}x=\infty$, and $\lim_{x\to\infty}\frac x{x^2}=\frac\infty\infty$, then one may apply L'H. – 2017-02-10
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    Yes you may, but you first need to show that the limit $\lim_{x\rightarrow \infty} \frac{1}{2x}$ exists prior to doing so. Only then you can formally claim the original limit is $0$. – 2017-02-10
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    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. – 2017-02-10
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    @SimplyBeautifulArt: L'Hospital concludes that _if_ the limit of $\frac{f'(x)}{g'(x)}$ exists, _then_ it equals the original limit, but you're missing the premise here. – 2017-02-10
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    @Simply Beautiful Art Please look up what L'Hospital Rule states. – 2017-02-10
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    http://mathworld.wolfram.com/LHospitalsRule.html – 2017-02-10
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    ^ Sorry if we use "original limit exists" differently, as I usually don't include the case that $\lim\frac{f(x)}{g(x)}=\pm\infty$ counts as "limit exists" like the definition says. – 2017-02-10
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    @HenningMakholm I honestly can't see what I'm missing. If you look through all of it, L'Hospital's rule was used correctly (in terms of how Wolfram says L'H is allowed to be used) but that is perhaps not the algebraic step that went wrong. Even in my answer L'H was used correctly... – 2017-02-10
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    @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. – 2017-02-10
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    Please look at the definition you sent. To apply a theorem, you must first make sure all of the conditions are satisfied. In our context, we have to first show that $\lim_{x\rightarrow\infty} \frac{1}{2x}=0$ and only then claim $\lim_{x\rightarrow\infty} \frac{x}{x^2}=0$, as it is a condition (see the word if at the end of the first line?) – 2017-02-10
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    @HenningMakholm How do you define "limit exists". – 2017-02-10
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    But even in the wikipedia article ..It is mentioned that it is assumed that the derivative ratio exists...You can look it up...As I cant hyperlink it – 2017-02-10
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    @Simply Beautiful Art This doesn't matter. In L'Hospital's case we can refer to $\pm \infty$ aswell. – 2017-02-10
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    @SimplyBeautifulArt: By the $\varepsilon$-$N$ definition, like in almost every introductory analysis text. – 2017-02-10
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    @HenningMakholm I am unsure what you mean "I'm ignoring the third premise". Could you be more specific? (sorry for taking your time) – 2017-02-10
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    @user35508 If you assume the derivative's ratio exists in the first hand there's no point to this problem because it's almost the same as the limit you need to calculate. – 2017-02-10
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    But I believe Proving something exists and finding a definite value for it are two different things..If one finds a definite value then surely so it means it also exists...(Sorry for bad English) – 2017-02-10
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    @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. – 2017-02-10
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    @HenningMakholm I see what your saying now. Thank you for the clarifications :-) And so there are two basic issues with this: the limit must "exist", and as my answer provides, arithmetic with some things doesn't quite work. – 2017-02-10
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    @user35508 These are indeed different things. But here, you are requested to prove that the limits exists AND it is 0. While applying L'Hospital's Rule, you already assumed that it exists without proving so. Therefore, your proof is flawed. The thing that you did prove is that if the limit exists, then it must be $0$, but this is different than the original problem. – 2017-02-10
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    @Theorem Even if the limit did "exist", it needn't need be $0$. It could also be $\pm\infty$. – 2017-02-10
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    @HenningMakholm..You said the RHS does not converge at all.. But doesn't as x approaches infinity the value must lie between the infimum and supremum of derivative ratio which is indeed true – 2017-02-10
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    @Simply Beautiful Art It could also be $\frac{\pi^2}{6}$. I don't see your point? – 2017-02-10
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    If the limit exists and is **finite**, it must be $0$. But my answer asks "and if we don't check to see if a limit is finite..." – 2017-02-10
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    Yes, assuming also that it is finite, which just proves more how flawed the solution is I guess. – 2017-02-10
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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.

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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 $\lim_{x\to0}\frac{\arcsin x -x}{x^2}$

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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 $(-\epsilon, \epsilon)$.

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Your expression In other words:

As x approaches infinity, then 1/x approaches 0 so its answer is 0 Try to think in that way...

Your method is wrong as you can only use L'Hospital's Rule when you know the limit of the derivative ratio exists.