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

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