Limit $\sin(x)\cdot \ln(\cos(x))$ as $x\to \pi/2$

Question

I just did a question on a test, that I unfortunately know now that I was unable to do correctly. For me the real bothersome part is not that I didn't do it correctly, but more that I don't understand how to do it.

$$\lim_{x\to\pi/2} \sin(x)\cdot \ln(\cos(x))$$

The question asked us to find it from both the left and the right. I looked it up on wolfrm alpha, which reference the power rule of limits, which we haven't learned so I don't imagine we were suppose to come to our conclusion that way. Either way though, evaluating the limit as is, I always get $1 \cdot{-\infty}$. We have learned that this is undefined though, so I am confused as to how the evaluation works. It seems from time to time people just do the multiplication and say $-\infty$ is the result.

My questions are as follows then:

• How can I go about understanding how this limit is evaluated?
• How would I be able to see that the limit only exists from the left without a visual aid?
• Since the limit only exists from the left, why is it possible to use it in an improper integral like:

$$\int_{0}^{\pi/2} \sin(x)\cdot \ln(\cos(x))\, dx$$

Answer 1

As $x\to\pi/2$ from the left, $\cos x$ goes downward to $0$, so $\ln\cos x$ goes downward to $-\infty$. But as $x\to\pi/2$ from the right, you'd be taking the logarithm of a negative number, and that is undefined.

You need to recall from trigonometry that $\cos x>0$ immediately to the left of $\pi/2$ and $\cos x<0$ immediately to the right of $\pi/2$.

Confusing language exists: One says that as $x\uparrow\pi/2$, then $\cos x$ "diverges" to $-\infty$ rather than $\cos x$ "converges" to $-\infty$, and that the limit does not exist, and that the limit is $-\infty$. To say that the limit exists is often taken to mean the limit is a real number and not one of $\pm\infty$. Theorems assert things like the limit of $f+g$ is the limit of $f$ plus the limit of $g$ provided both limits "exist", but the theorem does not apply if either limit is one of $\pm\infty$, since then the limit "does not exist".

However, as $x\downarrow\pi/2$, nothing exists whose limit is taken; it's not just the limit that is said not to exist; it's the values of the function.

The integral $\displaystyle\int_0^{\pi/2}$ does not depend on what happens when $x>\pi/2$, but only on what happens when $0<x<\pi/2$.

Answer 2

Operations on infinity are undefined, but when you are in a situation like $$ \lim_{x\to c}f(x)=l,\qquad \lim_{x\to c}g(x)=\infty $$ with $l>0$ (and finite), then you can surely say that $$ \lim_{x\to c}f(x)g(x)=\infty $$ Fix $M>0$; by hypothesis you can find

1. $\delta_1>0$ such that, for $0<|x-c|<\delta_1$, $g(x)>2M/l$
2. $\delta_2>0$ such that, for $0<|x-c|<\delta_2$, $f(x)>l/2$

Therefore, for $0<|x-c|<\delta=\min(\delta_1,\delta_2)$, we have $$ f(x)g(x)>\frac{2M}{l}\frac{l}{2}=M $$

Adapt to the cases $l<0$ and $-\infty$ and to the case of one-sided limits.

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Of course your limit can only be taken “from the left”, where $\cos x>0$. It depends on conventions whether you can talk about $$ \lim_{x\to\pi/2}\sin x\ln\cos x $$ or just about $$ \lim_{x\to\pi/2^-}\sin x\ln\cos x $$ (In my convention, I use the first notation.)

In any case, your improper integral can be written $$ \lim_{a\to\pi/2^-}\int_0^a\sin x\ln\cos x\,dx $$ and the function under the integral sign is defined in all intervals of the form $[0,a]$, for $0<a<\pi/2$. So it makes sense to ask about its convergence.