Limit problem, how to show algebraically this limit doesn't exist?

Question

I believe that the following limit does not exist: $$\lim\limits_{x \rightarrow \infty} \dfrac{\ln(1+\sin x)}{x} $$ A graphing tool suggests that there are vertical asymptotes at "multiples" of 3$\pi$/2. But I am not sure how to show this algebraically. It doesn't qualify for L'Hospital's Rule and series didn't get me very far either. Any suggestions are welcome. EDIT: I tried a u-sub by setting $1+sinx=t$. The expression becomes $\frac{\ln t}{\arcsin(t-1)}$ where $t$ is in the interval $[0,2]$. Now for all $t$ in the interval $(0,2]$, the limit exists (including $t=1$ which is just a basic L'Hospital Rule). For $t=0+$, it is clear the limit does not exist. That would be my best answer. EDIT 2: There is a problem with my answer. $1+sinx=t$ is not equivalent to $x=arcsin(t-1)$ because the arcsine gives a restricted outcome for $x$. It is not possible for $x$ to go to infinity assuming the expression $x=arcsin(t-1)$ in the expression's denominator.

Answer 1

As noticed the function is not defined for $x=\frac{3\pi}2+2k\pi$.

Excluding these points from the domain, the limit doesn't exist indeed by

• $x_n=\frac{3\pi}2+\frac1n+2n\pi$

we have that

• $\sin x_n =-\cos \left(\frac 1 n\right)=-1+\frac1{2n^2}+o\left(\frac1{n^2}\right)$

and then

$$\dfrac{\ln(1+\sin x_n)}{x_n}=\dfrac{\ln\left(\frac1{2n^2}+o\left(\frac1{n^2}\right)\right)}{\frac{3\pi}2+\frac1n+2n\pi}\to 0$$

but by

• $x_n=\frac{3\pi}2+\frac1{e^n}+2n\pi$

we have that

• $\sin x_n =-\cos \left(\frac 1 {e^n}\right)=-1+\frac1{2e^{2n}}+o\left(\frac1{e^{2n}}\right)$

and

$$\dfrac{\ln(1+\sin x_n)}{x_n}=\dfrac{\ln\left(\frac1{2e^{2n}}+o\left(\frac1{e^{2n}}\right)\right)}{\frac{3\pi}2+\frac1{e^n}+2n\pi}\to -\frac 1{\pi}$$

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Edit

In order to make things simpler for students, we can proceed also by l'Hospital as follows.

For the first path let $t=\frac1n \to 0^+$ then

• $x_n=\frac{3\pi}2+\frac1n+2n\pi=\frac{3\pi}2+t+\frac{2\pi}t$
• $\sin x_n =-\cos t$

and then

$$\lim_{n\to \infty}\dfrac{\ln(1+\sin x_n)}{x_n} =\lim_{t\to 0^+}\dfrac{\ln\left(1-\cos t\right)}{\frac{3\pi}2+t+\frac{2\pi}t} \stackrel{H.R.}=\lim_{t\to 0^+}\dfrac{\sin t}{\left(1-\cos t\right)\left(1-\frac{2\pi}{t^2}\right)}=\\ =\lim_{t\to 0^+}\dfrac{t^2\sin t}{\left(1-\cos t\right)\left(t^2-2\pi\right)}=0$$

For the second path let $t=\frac1{e^n} \to 0^+$ then

• $x_n=\frac{3\pi}2+\frac1{e^n}+2n\pi=\frac{3\pi}2+t-2\pi\log t$
• $\sin x_n =-\cos t$

and then

$$\lim_{n\to \infty}\dfrac{\ln(1+\sin x_n)}{x_n} =\lim_{t\to 0^+}\dfrac{\ln\left(1-\cos t\right)}{\frac{3\pi}2+t+2\pi\log t} \stackrel{H.R.}=\lim_{t\to 0^+}\dfrac{\sin t}{\left(1-\cos t\right)\left(1-\frac{2\pi}t\right)}=\\ =\lim_{t\to 0^+}\dfrac{\sin t}{t}\dfrac{t^2}{\left(1-\cos t\right)\left(t-2\pi\right)}=-\frac 1 \pi$$

Answer 2

@user 's answer is extremely nice, but I want to include my own answer, which relies on the definition of a limit, because it gives additional insight into how limits are rigorously defined.

Let's consider for a moment what $\lim\limits_{x \rightarrow \infty} f(x)$ actually means.

By the definition of a limit, for $\lim\limits_{x \rightarrow \infty} f(x) = C$, where C is a finite number, then for all $\epsilon > 0$, there exists an $M$ such that for all $x > M, |f(x)-L| < \epsilon$. This is the formal mathematics way of saying that "a limit of $f(x)$ for $x$ 'at' infinity equalling $L$ really means that no matter how close you want to get to $L$, you can ALWAYS find some number $M$ such that all $x > M$, $f(x)$ is at LEAST that close to $L$."

On the other hand, for $\lim\limits_{x \rightarrow \infty} f(x) = \pm\infty$, then for any $r \in \mathbb{R}$, there exists an $M$ such that for all $x > M, f(x) > r$ (for $+ \infty$) or $f(x) < r$ (for $- \infty$). his is the formal mathematics way of saying that "a limit of $f(x)$ for $x$ 'at' infinity equalling $\infty$/$-\infty$ really means that no matter how how big/small you want your number to be, you can ALWAYS find some number $M$ such that all $x > M$, $f(x)$ is ALWAYS at LEAST that big/small."

So what does that mean for this limit?

Consider for a moment the behavior of $\ln(1+\sin(x))$. $1+\sin(x)$ is periodic, and is always $[0,2]$, always hitting $0$ at $\frac{3\pi}{2}k$, where $k$ is an integer. Importantly, no matter what $x$ is, $1+\sin(x)$ always approaches $0$ from POSITIVE values. It is known that, as the positive argument logarithm gets arbitrarily close to $0$, the logarithm gets as small as you want. We express this fact as $\lim\limits_{x \rightarrow 0^+} \ln(x) = -\infty$.

Therefore, for all x, $\ln(1+\sin(x))$ gets arbitrarily small over a period of $2\pi$. Those are the asymptotes you're seeing.

But $\ln(1+\sin(x))$ always hits positive values over a period of $2\pi$. Now, for $f(x) = \frac{\ln(1+\sin(x))}{x}$, this function ALSO hits positive values over a period of $2\pi$, but these positive values keep getting closer to zero. BUT, because of the way the numerator works, these positive values are bounded below by 0. So, over every $2\pi$ period, $f(x)$ gets smaller than any value you want AND always gets bigger than zero.

Therefore, no matter the value for $x$ (say $x = M$), there's no finite $L$ such that for all $x > M$, $f(x)$ is within some small distance of $L$, because $f(x)$ always gets arbitrarily small over every period of $2\pi$, so the limit cannot be finite. But we also know that the limit can't be $-\infty$, because there's no $M$ such that for all $x > M$, $f(x)$ is always smaller than any negative number, because $f(x)$ always ends up bigger than $0$ over every period of $2\pi$. And the function does not increase without bound (I leave it as an exercise for you to figure out the proof for this). So the limit cannot be finite and cannot be infinite. Therefore, it does not exist.