$\lim_{x \to + \infty} \left \{ x-x^2 \ln (1+ \sin \frac{1}{x}) \right \}$

Question

I've tried to solve this limit: $\lim_{x \to + \infty} \left \{ x-x^2 \ln (1+ \sin \frac{1}{x}) \right \} $

$x-x^2 \ln (1+ \sin \frac{1}{x}) \sim x-x^2 \sin \frac{1}{x} \sim x-x = 0 $

but the final result should be $\frac{1}{2}$

Answer 1

Set $\dfrac1x=h$ to find

$$\lim_{h\to0^+}\dfrac{h-\ln(1+\sin h)}{h^2}$$

$$=\lim_{h\to0^+}\dfrac{h-\sin h+\dfrac{\sin^2h}2+O(h^3)}{h^2}$$

$$=\lim_{h\to0^+}\dfrac{h-\sin h}{h^2}+\dfrac12\left(\lim_{h\to0^+}\dfrac{\sin h}h\right)^2+\lim_{h\to0^+}\dfrac{O(h^3)}{h^2}$$

For the first part, use the second problem of Are all limits solvable without L'Hôpital Rule or Series Expansion

Alternatively,

$$\lim_{h\to0^+}\dfrac{h-\ln(1+\sin h)}{h^2}=\lim_{h\to0^+}\dfrac{h-\sin h}{h^2}+\left(\lim_{h\to0^+}\dfrac{\sin h}h\right)^2\cdot\lim_{h\to0^+}\dfrac{\sin h-\ln(1+\sin h)}{\sin^2h}$$

For the last part, use $(3)$ problem of Are all limits solvable without L'Hôpital Rule or Series Expansion

Answer 2

$$\lim x-x^2 \ln (1+ \sin \frac{1}{x})\\=\lim x-x^2\left(\sin \frac{1}{x}-\frac 1 2 \sin^2 \frac{1}{x}+\frac1 3 \sin^3 \frac{1}{x}+\cdots\right)\\= \lim x-x^2 \sin \frac{1}{x} +\frac1 2\\=\lim x-x^2\left(\frac 1 x-\frac 1{6x^3}+\frac 1 {120x^5}-\cdots\right)+\frac 1 2\\=\frac 1 2$$.

I just used expansion of $\ln(1+x),\sin (x)$ and the fact $\lim_{x\to 0}\frac {\sin x}{x}=1$

Answer 3

Convert this limit by substituting $$x=\frac{1}{t}$$ as $x \rightarrow +\infty, t \rightarrow 0$ therefore, $$\lim_{x \to + \infty} \left \{ x-x^2 \ln (1+ \sin \frac{1}{x}) \right \} = \lim_{t \to 0} \left \{ \frac{1}{t}-\frac{1}{t^2} \ln (1+ \sin t) \right \} $$ $$\lim_{t \to 0} \left \{ \frac{t - \ln (1+ \sin t)}{{t^2}} \right \} =\lim_{t \to 0} \left \{ \frac{t - (\sin t - \frac{\sin^2t}{2} + \frac{\sin^3t}{3} -\frac{sin^4t}{4}...)}{{t^2}} \right \} $$ $$\lim_{t \to 0} \left \{ \frac{t - \sin t + \frac{\sin^2t}{2} - \frac{\sin^3t}{3} +\frac{sin^4t}{4}...}{{t^2}} \right \} =\lim_{t \to 0} \left \{ \frac{t - \sin t}{t^2} + \frac{\sin^2t}{2t^2}+\frac{ - \frac{\sin^3t}{3} +\frac{sin^4t}{4}...}{{t^2}} \right \} $$ $$\lim_{t \to 0}{\frac{t-\sin t}{t^2}}= \lim_{t \to 0}\frac{t-(t-\frac{t^3}{3!}-\frac{t^5}{5!}...)}{t^2}=0$$ $$\lim_{t \to 0}{\frac{\sin^2t}{2t^2}=\lim_{t \to 0}\frac{1}{2}(\frac{\sin t}{t})^2}=\frac{1}{2}$$ $$\lim_{t \to 0}{\frac{-\frac{\sin^3t}{3!}+\frac{\sin^5t}{5!}...}{t^2}=0}$$ So you get: $$\lim_{t \to 0} \left \{ \frac{t - \ln (1+ \sin t)}{{t^2}} \right \}=0+\frac{1}{2}+0 = \frac{1}{2}$$ Hope this helps...

Answer 4

In the following proof, I won't be using L'hopital or series expansion.

Let $ x\in\mathbb{R}^{*} $, we have :

\begin{aligned}x-x^{2}\ln{\left(1+\sin{\left(\frac{1}{x}\right)}\right)}=\frac{\frac{1}{x}-\sin{\left(\frac{1}{x}\right)}}{\frac{1}{x^{2}}}+\left(x\sin{\left(\frac{1}{x}\right)}\right)^{2}\left(\frac{\sin{\left(\frac{1}{x}\right)}-\ln{\left(1+\sin{\left(\frac{1}{x}\right)}\right)}}{\sin^{2}{\left(\frac{1}{x}\right)}}\right)\end{aligned}

We ended with well known limits, that I'm sure have been computed many times on this site.

$ \textbf{Let's proof first that : $ \displaystyle\lim_{x\to 0}{\frac{x-\sin{x}}{x^{2}}}=0$ : } $

Note that for any $ x\in\mathbb{R}^{*} $, we have : $$ \frac{x-\sin{x}}{x^{2}}=\int_{0}^{1}{\left(1-y\right)\sin{\left(xy\right)}\,\mathrm{d}y} $$

Thus : \begin{aligned} \left|\frac{x-\sin{x}}{x^{2}}\right|&\leq\int_{0}^{1}{\left(1-y\right)\left|\sin{\left(xy\right)}\right|\mathrm{d}y}\\ &\leq\int_{0}^{1}{\left(1-y\right)\left|xy\right|\mathrm{d}y}\\ \left|\frac{x-\sin{x}}{x^{2}}\right|&\leq\frac{\left|x\right|}{6} \end{aligned}

Hence : $$ \lim_{x\to 0}{\frac{x-\sin{x}}{x^{2}}}=0 $$

$\textbf{Let's proof then that : $ \displaystyle\lim_{x\to 0}{\frac{x-\ln{\left(1+x\right)}}{x^{2}}}=\frac{1}{2} $ : }$

Note that for any $ x\in\left(-1,1\right)\setminus\left\lbrace 0\right\rbrace $, we have : $$ \frac{x-\ln{\left(1+x\right)}}{x^{2}}=\int_{0}^{1}{\frac{1-y}{\left(1+xy\right)^{2}}\,\mathrm{d}y}=\frac{1}{2}-x\int_{0}^{1}{\frac{y\left(2+xy\right)}{\left(1+xy\right)^{2}}\,\mathrm{d}y} $$

Since : \begin{aligned} \left|x\int_{0}^{1}{\frac{y\left(2+xy\right)}{\left(1+xy\right)^{2}}\,\mathrm{d}y}\right|&\leq\left|x\right|\int_{0}^{1}{\frac{y\left(2+\left|x\right|y\right)}{\left(1-\left|x\right|y\right)^{2}}\,\mathrm{d}y}\\ &\leq\frac{\left|x\right|\left(2+\left|x\right|\right)}{2\left(1-\left|x\right|\right)^{2}}\underset{x\to 0}{\longrightarrow}0 \end{aligned}

we get that : $$ \lim_{x\to 0}{\frac{x-\ln{\left(1+x\right)}}{x^{2}}}=\frac{1}{2} $$

$\textbf{Now let's apply this to our expression : }$

\begin{aligned}\lim_{x\to +\infty}{\left(x-x^{2}\ln{\left(1+\sin{\left(\frac{1}{x}\right)}\right)}\right)}&=\lim_{x\to +\infty}{\left(\frac{\frac{1}{x}-\sin{\left(\frac{1}{x}\right)}}{\frac{1}{x^{2}}}+\left(x\sin{\left(\frac{1}{x}\right)}\right)^{2}\left(\frac{\sin{\left(\frac{1}{x}\right)}-\ln{\left(1+\sin{\left(\frac{1}{x}\right)}\right)}}{\sin^{2}{\left(\frac{1}{x}\right)}}\right)\right)}\\ &=0 + 1^{2}\times\frac{1}{2}\\ &=\frac{1}{2}\end{aligned}