L'hopital rule fails with limits to infinity?

Question

$$ \lim_{n \to \infty} \frac{1 +cn^2}{(2n+3 + 2 \sin n)^2} = ? $$

if I factor the $n^2$ out of denominator,

$$ \lim_{n \to \infty} \frac{ 1 + cn^2}{ n^2 ( 2 + 3n^{-1} + 2 \frac{ \sin n}{n} )^2}$$

And take limit directly, I get the answer as

$$ \frac{c}{4}$$

However, If I apply l'hopital rule, Iget

$$ \lim_{ n \to \infty} \frac{ 2cn}{2 (2n + 3 + 2 \sin n)( 2 + 2 \cos n)} $$

However this new limit gives a different value than original according to wolfram.. and neither am I able to compute it by hand, what am I missing?

Some people say of limit existing and not existing, but then suppose

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

Does this limit exist? how do you define a limit to be existing as in what is sufficent condition for it

As far as I know , I've been taught that I can apply for 0/0 and infinity/ infinity forms — tryst with freedom, Jul 23 '20 at 22:08
@DDD4C4U One can apply LHR to quotients on the form $\frac{\text{ANYTHING}}{\infty}$ — Mark Viola, Jul 23 '20 at 22:09
@DDD4C4U Since the limit of the resulting (i.e. differentiated) denominator fails to exist, LHR is simply inapplicable here. — Mark Viola, Jul 23 '20 at 22:18
what does it mean to fail to exist? the 'n' term beats the growth of sine term so I think it's not totally undefined — tryst with freedom, Jul 23 '20 at 22:19
the problem that I Have precisely is that does limit being infinity mean that is defined? because I think 1/x limit as x go to 0 is not. And in that case we can take l'hopital but not here — tryst with freedom, Jul 23 '20 at 22:25
This is the same thing (conceptually) with $\frac{n}{n+\sin n}$. — , Jul 23 '20 at 22:27
Lhopital is not applicable here since the limit of the expeossion obtained after its fist application does not exist. L’Hospital says that under 0/0 or $\infty/\infty$ conditions, if the ratio of derivative exists then the limit of the original ratio exists. It DOES NOT say that if the limit of the ratios under the above indeterminacies exists, then the ratio of derivatives exists. — Mittens, Jul 23 '20 at 22:47
This old answer of mine may help explain things: https://math.stackexchange.com/questions/1710786/why-does-lhopitals-rule-fail-in-calculating-lim-x-to-infty-fracxx-s/1710798#1710798 — Barry Cipra, Jul 23 '20 at 23:37
@DDD4C4U you say the wrong here, $ \lim_{x \to 0} \frac{1}{x} = \infty$, it doesn't exist infinitely also! — A learner, Jul 24 '20 at 09:18
@OliverDiaz To be more precise, LHR rule is applicable to a quotient of the form $\frac{\text{ANYTHING}}{\infty}$. The limit of the numerator need not even exist provided the limit of the denominator is $\infty$. — Mark Viola, Jul 24 '20 at 14:08

score 4 · Accepted Answer · answered Jul 23 '20 at 22:39

4

The rule of L'Hospital states that the limit of $\dfrac fg$ equals that of $\dfrac{f'}{g'}$ if the latter exists. You precisely found a case where this does not hold.

We can simplify the example as

$$\lim_{n\to\infty}\frac{n+\sin n}n=1$$

but

$$\lim_{n\to\infty}\frac{1+\cos n}1$$ is undefined.

answered Jul 23 '20 at 22:39

2

This is not what L'Hôpital's rule says. There are lots of cases where the limit of $f^{\prime}/g^{\prime}$ exists and is not equal to the limit of $f/g$. The rule only applies to specific indeterminate forms (and the example given is not one of these forms). I know that you know this, but your answer is misleading by excluding it. – xxxxxxxxx Jul 24 '20 at 05:19
@MorganRodgers: this is clearly a $\dfrac{\infty}{\infty}$ form, the limit of which is $\dfrac c4$. What do I "exclude" ? – Jul 24 '20 at 06:18
@MorganRodgers I'm interested in the answer you would like to give for this – tryst with freedom Jul 24 '20 at 08:55
what does it mean for a limit to exist? is a limit like infinity/ infinity said to exist? I don't think I fully understood l'hopitals as I"ve never seen the proof for lims at inf – tryst with freedom Jul 24 '20 at 08:56
@DDD4C4U: it is the purpose of limits calculus to determine if a limit exists or not. – Jul 24 '20 at 11:13
@DDD4C4U If this helps, the reason that the limit in this answer does not exist is that cosine oscillates forever. You can't plug in a really big value for $n$ and see the function approaching anything, because it doesn't approach anything - it just goes up and down. That's why we say the limit does not exist in this case. – Polygon Jul 24 '20 at 13:48
so hwo does oscillating limits relate to limit existing or not? – tryst with freedom Jul 25 '20 at 09:26
@DDD4C4U: what is $\lim_{x\to \infty}\cos x$ in your opinion ? – Jul 25 '20 at 09:28
-1 or 0 or 1 , the average would be 0. Jokes aside, I get that we cant say oscillating limits, but then what does it mean for limit to exist? – tryst with freedom Jul 25 '20 at 09:30
@DDD4C4U: https://en.wikipedia.org/wiki/Limit_(mathematics) – Jul 25 '20 at 10:06
I think the important question that hasn't been answered above (and which I personally also don't know the answer to) is whether L'Hôpital's rule applies when the limit of the quotient of the derivatives is infinite (i.e. the rule only applies when the limit of the quotient of the derivatives exists, but does "the limit existing" include or exclude infinite limits? Irrelevant for this question, as in this case the limit does not exist at all, but I'm still curious about the case of an infinite limit). – Tom Oct 03 '22 at 18:41

L'hopital rule fails with limits to infinity?

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