L'Hopital's rule fails with limits to infinity?

Question

Let consider

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

If I factor the $n^2$ out of denominator

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

and take limit directly, I get the answer as

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

However, if I apply l'hopital rule, I get

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

However this new limit gives a different value than original according to wolfram and neither I am 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?

Answer 1

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.

Answer 2

The case is inapplicable because of one of the conditions to use L’Hopital rule fails, that is,

$$g’(x) \ne 0, \forall x \text{ around } c$$

In this case, the resulting denominator will be zero if $2 + 2\cos{x}=0$, which, in fact, has infinitely many real roots.

Therefore, the denominator can be zero around “infinity”, hence the case failed.

For further clarification, visit https://en.m.wikipedia.org/wiki/L'H%C3%B4pital's_rule, the 3rd counterexample is what I’m demonstrating.

Hope this helps!

Answer 3

As noticed L’Hopital's rule doesn't work in this case but we can say something more on that.

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Let consider at first the case $n\in \mathbb R$.

As already discussed here, one way to conlude that L’Hopital's rule doesn't work is by the hypotesis adopted in the theorem which states that for the denominator $g'(x)\neq 0$ in the interval.

But what if we exclude such values from the domain? that is, if we assume

$$2+2\cos n \neq 0 \iff n\neq \pi +2k\pi$$

Well, in this case also L’Hopital's rule doesn't work too but for another reason that is that the new limit

$$\lim_{ n \to \infty} \frac{ 2cn}{2 (2n + 3 + 2 \sin n)( 2 + 2 \cos n)}=\lim_{ n \to \infty} \frac{ c}{ \left(2 + \frac3n + 2 \frac{\sin n}n\right)}\frac1{ 2 + 2 \cos n}$$

doesn't exist, indeed

$$\frac{ c}{ \left(2 + \frac3n + 2 \frac{\sin n}n\right)}\to \frac c 4$$

but

$$\lim_{ n \to \infty}\frac1{ 2 + 2 \cos n}$$

is problematic indeed

• for $x_n=2kn \to \infty \implies \frac1{ 2 + 2 \cos n}\to \frac14$
• for $x_n=\frac \pi 2+2kn \to \infty \implies \frac1{ 2 + 2 \cos n}\to \frac12$

such that the new limit doesn't exist and then we again can't conclude nothing by l'Hospital rule.

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When $n\in \mathbb Z$ again L’Hopital's rule doesn't work because the limit for $\cos n$ doesn't exists, as discusse here

• Prove the divergence of the sequence $\left\{ \sin(n) \right\}_{n=1}^{\infty}$.

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The first way you have used is the most clear and effective, but if we want apply l'Hospital rule, a way is passing by squeeze theorem

$$ \frac{1 +cn^2}{(2n+5)^2}\le \frac{1 +cn^2}{(2n+3 + 2 \sin n)^2}\le \frac{1 +cn^2}{(2n+1)^2}$$

and then applying the rule for the bounds I don't suggest this way, but in this case the application is formally fine).

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For the last question we usually say

• limit exists (finite): $\lim_{x\to x_0} f(x) = L\in \mathbb R$
• limit exists (not finite, diverging cases): $\lim_{x\to x_0} f(x) = \infty$ or $\lim_{x\to x_0} f(x) = -\infty$
• limit doesn't exist, otherwise (e.g. $\sin x$, $(-1)^n$, $\cos n$, etc.)

The limit in the last example doesn't exist since $\lim_{x \to 0^+} \frac{1}{x} = \infty$ but $\lim_{x \to 0^-} \frac{1}{x} = -\infty$.