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

## Main Question or Discussion Point

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

$$\lim_{x\to 0} \frac{\ln (1-x) - \sin x}{1 - \cos^2 x}$$

does not exist?

tiny-tim
Homework Helper
l'Hôpital?

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

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.

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

$$\lim_{x\to 0} \frac{\ln (1-x) - \sin x}{1 - \cos^2 x}$$

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

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

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

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.)

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

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

$$\lim_{x\to a} \Big( f(x) + g(x) \Big)$$

by considering

$$\lim_{x\to a} f(x) + \lim_{x\to a} g(x),$$

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

tiny-tim
Homework Helper
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

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
Homework Helper
$$\lim_{x\to 0} x \frac{\ln (1-x) - \sin x}{1 - \cos^2 x}=-2$$

D H
Staff Emeritus
$$\lim_{x\to 0} x \frac{\ln (1-x) - \sin x}{1 - \cos^2 x}=-2$$
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.

You seem to be assuming that we can determine

$$\lim_{x\to a} \Big( f(x) + g(x) \Big)$$

by considering

$$\lim_{x\to a} f(x) + \lim_{x\to a} g(x),$$

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.