Does anyone know the best way to show this limit does not exist?

[AxiomOfChoice]

What's the most efficient way (other than just plotting it and eyeballing it) to show that

<br /> \lim_{x\to 0} \frac{\ln (1-x) - \sin x}{1 - \cos^2 x}<br />

does not exist?

 

[tiny-tim]

l'Hôpital?

 

[DonAntonio]

l'Hôpital?

Can't be. L'Hospital gives a sufficient condition for a limit to exist, not necessary, meaning: if the limit *after* doing L'H doesn't exist we cannot deduce the original limit doesn't exist either.

DonAntonio

 

[micromass]

Can't be. L'Hospital gives a sufficient condition for a limit to exist, not necessary, meaning: if the limit *after* doing L'H doesn't exist we cannot deduce the original limit doesn't exist either.

DonAntonio

Still, we can use L'Hopitals rule by noting it holds also for one-sided limits and for extended limits. So if we can show that the left-sides and right-sides limits of the derivatives are infinite, then so are the left-sided and rigt-sided limits of the original function.

 

[DonAntonio]

What's the most efficient way (other than just plotting it and eyeballing it) to show that

<br /> \lim_{x\to 0} \frac{\ln (1-x) - \sin x}{1 - \cos^2 x}<br />

does not exist?

One way is using Taylor expansions: \log(1-x)=-(x+\frac{x^2}{2}+\frac{x^3}{3}+...)\,\,,\,\,\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-...\Longrightarrow \sin^2x=x^2-\frac{x^4}{3}+... so\frac{\log(1-x)\sin x}{\sin^2x}=\frac{-2x-\frac{x^2}{2}-\frac{x^3}{6}-...}{x^2-\frac{x^4}{3}+...}=-\frac{1}{x}\frac{2+\frac{x}{2}+\frac{x^2}{3}+...}{1-\frac{x^2}{3}+...}

The second factor in the last expression above has limit \,2\,\text{when}\,x\to 0\, , whereas the first factor has no limit, as it approaches \,-\infty\,\,\,or\,\,\,+\infty\, according as whether \,x\to 0^+\,\,\,or\,\,\,x\to 0^-\, , resp.

DonAntonio

 

[Bohrok]

\frac{\ln(1-x)-\sin x}{1-\cos^2x} = \frac{\ln(1-x)-\sin x}{\sin^2x} = \frac{\ln(1-x)}{\sin^2x} - \csc x
\lim_{x\to 0^-} \frac{\ln(1-x)}{\sin^2x} = +\infty \lim_{x\to 0^+}\frac{\ln(1-x)}{\sin^2x} = -\infty\lim_{x\to 0^-}-\csc x = +\infty\lim_{x\to 0^+}-\csc x = -\infty

The left and right-sided limits aren't the same, so the limit doesn't exist.

 

[AxiomOfChoice]

Still, we can use L'Hopitals rule by noting it holds also for one-sided limits and for extended limits. So if we can show that the left-sides and right-sides limits of the derivatives are infinite, then so are the left-sided and rigt-sided limits of the original function.

Does this mean that L'Hopital's rule is still good even if we rewrite the statement of the theorem in terms of left-hand (or right-hand) limits? (We obviously then just obtain a result about the left-hand or right-hand limit of the function.)

 

[AxiomOfChoice]

\frac{\ln(1-x)-\sin x}{1-\cos^2x} = \frac{\ln(1-x)-\sin x}{\sin^2x} = \frac{\ln(1-x)}{\sin^2x} - \csc x
\lim_{x\to 0^-} \frac{\ln(1-x)}{\sin^2x} = +\infty \lim_{x\to 0^+}\frac{\ln(1-x)}{\sin^2x} = -\infty\lim_{x\to 0^-}-\csc x = +\infty\lim_{x\to 0^+}-\csc x = -\infty

The left and right-sided limits aren't the same, so the limit doesn't exist.

You seem to be assuming that we can determine

<br /> \lim_{x\to a} \Big( f(x) + g(x) \Big)<br />

by considering

<br /> \lim_{x\to a} f(x) + \lim_{x\to a} g(x),<br />

even if the limits of f and g are \pm \infty. Is this really the case?

 

[tiny-tim]

Does this mean that L'Hopital's rule is still good even if we rewrite the statement of the theorem in terms of left-hand (or right-hand) limits?

yes … if we can use l'hôpital to prove that the left and right limits are ∞ and -∞, then we have proved that "the" limit does not exist

 

[TylerH]

Does this mean that L'Hopital's rule is still good even if we rewrite the statement of the theorem in terms of left-hand (or right-hand) limits? (We obviously then just obtain a result about the left-hand or right-hand limit of the function.)

A function has a limit at x0 iff its right and left limits at x0 exist and are equal.

 

[lurflurf]

<br /> \lim_{x\to 0} x \frac{\ln (1-x) - \sin x}{1 - \cos^2 x}=-2<br />

 

[D H]

<br /> \lim_{x\to 0} x \frac{\ln (1-x) - \sin x}{1 - \cos^2 x}=-2<br />

Nope. That's \lim_{x\to 0} x \frac{\ln (1-x) - \sin x}{\sin x}. It looks like you lost a power of 2 in the denominator.

 

[laughingebony]

You seem to be assuming that we can determine

<br /> \lim_{x\to a} \Big( f(x) + g(x) \Big)<br />

by considering

<br /> \lim_{x\to a} f(x) + \lim_{x\to a} g(x),<br />

even if the limits of f and g are \pm \infty. Is this really the case?

It depends on whether the infinite limits have the same sign. If, say, \lim_{x\to a} f(x) = \infty while \lim_{x\to a} g(x) = -\infty, then we have to use other methods to investigate the limit. But, informally speaking, if \lim_{x\to a} f(x) = \lim_{x\to a} g(x) = \infty (note: same sign), then close to a, f(x) is very large and g(x) is nonnegative, which means that f(x)+g(x)≥f(x)+0, which is still very large.