# Evaluate $\lim_{n\to\infty}\frac{1}{\sin(n)}-\frac{1}{n}$ and $\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{n-kx}$ using L'Hospital and Riemann sum

I have to calculate the two following limits:
a) $$\lim_{n\to\infty}\frac{1}{\sin(n)}-\frac{1}{n}$$.
b) $$\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{n-kx}$$ for $$-1

Hint : use L'Hospital and Riemann sums.

a) So first I get common denominator $$\lim_{n\to\infty}\frac{n-\sin(n)}{n\sin(n)}$$, then I use L'Hospital $$\lim_{n\to\infty}\frac{1-\cos(n)}{\sin(n)+n\cos(n)}$$. Now, if $$n$$ is not an odd multiple of $$\frac{\pi}{2}$$, we get $$0$$. If it's an odd multiple, we get $$\pm 1$$. Now I'm not sure about my method because WolframAlpha gets another result : https://www.wolframalpha.com/input/?i=limit+as+n+approaches+infinity+of+1%2Fsin(n)-1%2Fn. Their result is $$-\infty$$ to$$-1$$, $$1$$ to $$\infty$$

b) Here I thought about factoring out an $$\frac{1}{n}$$ and making the substitution $$x=\frac{k}{n}$$, so we get
$$\int_{0}^{1}\frac{1}{1-x^2}dx=\int_{0}^{1}\frac{1}{(1-x)(1+x)}dx$$. Now we could make partial fraction decomposition to get $$\frac{1}{2(x+1)}-\frac{1}{2(x-1)}$$. And so, if we integrate that, we get $$\frac{1}{2}\log(x+1)-\frac{1}{2}\log(x-1)$$. Now what I find strange is that first we can let $$x=\frac{k}{n}$$ if $$x$$ is already in the equation. Second, if we evaluate that from $$0$$ to $$1$$, we get $$\frac{1}{2}\log(2)+\frac{1}{2}\log(-1)-\frac{1}{2}\log(0)$$ so I don't know if my approach is correct.

Edit 2: For b), as said, we need to use another variable, so we get $$\int_0^1\frac{1}{1-yx}dy=-\frac{\log(1-x)}{x}$$ which seems valid if, as given $$-1. For a), as said, the limit does not exist. They probably meant the limit as n approaches 0. In this case, we can use l'Hospital a second time to get $$\frac{\sin(n)}{\cos(n)+\cos(n)-n\sin(n)}$$ which gives $$0$$ as n approaches zero.

• For part b): call $k/n=y$, you will see your mistake. – lcv Jan 17 '19 at 9:08

1. Technically speaking, it is illegal to use L'Hopital rule to sequential limits. And I don't think such limit exists. Since the hint is the L'Hopital rule, I think it is more likely to be $$\lim_{x \to 0} \frac 1{\sin x} - \frac 1x.$$ To let the limit be nonzero, maybe it also could be $$\lim_{x \to 0} \frac 1{\sin^2 x} - \frac 1{x^2}.$$
2. You got the letters wrong. $$x$$ is a given constant. To write Riemann sum you should consider the function $$f(t) = \frac 1{1 - x t}, t \in [0,1].$$

### UPDATE

If you insist, then such limit does not exist.

Proof. Assume such limit exists, let it be $$A$$, then using the arithmetic operation of limits, $$\lim_{n \to \infty} \frac 1{\sin n} = \lim_{n \to \infty} \frac 1{\sin n} - \frac 1n + \lim_{n \to \infty} \frac 1n = A + 0 = A.$$ Easy to see that $$A \neq 0$$ because $$\left\vert \frac 1 {\sin n} \right \vert \geqslant 1.$$ Then using the arithmetic operation again, $$\lim_{n\to \infty} \sin n = \frac 1A$$ exists. But in fact $$\sin n$$ has no limits [proof omitted, if you need then I will add], contradiction. Hence the limit does not exist. $$\square$$

• Well, I have this old exam in front of me and they clearly state $\lim_{n\to \infty}$ for a) I think I will add a picture in my question. – Frieder Jan 17 '19 at 9:16
• See edit with picture – Frieder Jan 17 '19 at 9:18
• Did they maybe make a typo in the exam and communicated it during the exam ? It's an exam from 2012 – Frieder Jan 17 '19 at 9:20
• @Poujh I don't know…… And I cannot possibly know. – xbh Jan 17 '19 at 9:32
• Yeah, I know, but I don't know too haha. I'm just making suppositions. – Frieder Jan 17 '19 at 9:33

a) This limit (obviously) does not exist. Are you sure it is correctly stated?

b) Your approach seems right, but if x is already in there, you need to call your dummy variable different, of course. As a consequence you will get an answer depending on x.

• a) Yes, I have this old exam question in front of me, and yes it is correctly stated. b) I will let you know what I get – Frieder Jan 17 '19 at 9:10
• For b), I get $\frac{-\log(yx-1)}{x}$ evaluated between 0 and 1, which seems again problematic – Frieder Jan 17 '19 at 9:15
• See my picture in the edit – Frieder Jan 17 '19 at 9:19
• Plug in $y = 0$ and $y=1$. Also you lost a minus sign somewhere, should be $\frac{-\log(1-x)}{x}$. For (a) they probably made a mistake, $n \to 0$ would make more sense. – Klaus Jan 17 '19 at 9:26
• Well that's why I said you lost a minus sign sowhere. ;-) – Klaus Jan 17 '19 at 9:28