How to compute $\lim\limits_{x \to +\infty} \left(\frac{x^4+x^5\sin\left(\frac{1}{x}\right)}{x^4\ln\left(\frac{x}{2x+1}\right)-x} \right)$? I have a problem with this limit, I don't know what method to use.
Can you show a method for the resolution ?
$$\lim\limits_{x \to +\infty} \left(\frac{x^4+x^5\sin\left(\frac{1}{x}\right)}{x^4\ln\left(\frac{x}{2x+1}\right)-x} \right)$$ Thanks
 A: $$\lim\limits_{x \to +\infty} \left(\frac{x^4+x^5\sin\left(\frac{1}{x}\right)}{x^4\ln\left(\frac{x}{2x+1}\right)-x} \right)=\lim\limits_{x \to +\infty} \left(\frac{1+x\sin\left(\frac{1}{x}\right)}{\ln\left(\frac{x}{2x+1}\right)-\frac{1}{x^3}} \right)$$
$$=\lim\limits_{x \to +\infty} \left(\frac{1+\frac{\sin\left(\frac{1}{x}\right)}{\frac{1}{x}}}{\ln\left(\frac{1}{2+\frac{1}{x}}\right)-\frac{1}{x^3}} \right)$$
So clearly $x \to \infty \Rightarrow\frac{1}{x}\to0$
So $$=\lim\limits_{x \to +\infty} \left(\frac{1+\frac{\sin\left(\frac{1}{x}\right)}{\frac{1}{x}}}{\ln\left(\frac{1}{2+\frac{1}{x}}\right)-\frac{1}{x^3}} \right)= \frac{1+1}{\ln\left(\frac{1}{2+0}\right)-0} $$
A: Let $\frac{1}{x}=t\implies t\to 0$ as $x\to \infty$, $$\lim_{x\to +\infty}\frac{x^4+x^5\sin\left(\frac 1x\right)}{x^4\ln\left(\frac{x}{2x+1}\right)-x}=\lim_{x\to +\infty}\frac{\frac 1x+\sin\left(\frac 1x\right)}{\frac 1x\ln\left(\frac{1}{2+\frac1x}\right)-\frac 1{x^4}}$$
$$=\lim_{t\to 0}\frac{t+\sin t}{-t\ln(2+t)-t^4}$$
$$=\lim_{t\to 0}\frac{1+\frac{\sin t}{t}}{-\ln(2+t)-t^3}$$
$$=\frac{1+1}{-\ln 2-0}=\color{red}{-\frac{2}{\ln 2}}$$
A: Note that we can write the relationships
$$\sin\left(\frac1x\right)=\frac1x+O\left(\frac1{x^3}\right) \tag 1$$
and 
$$\begin{align}
\log\left(\frac{x}{2x+1}\right)&=-\log 2+\log\left(1-\frac{1}{2x+1}\right)\\\\
&=-\log 2-O\left(\frac{1}{2x+1}\right) \tag 2
\end{align}$$
Using $(1)$ and $(2)$ reveals
$$\begin{align}
\frac{x^4+x^5\sin\left(\frac1x\right)}{x^4\log\left(\frac{x}{2x+1}\right)-x}&=\frac{2x^4+O\left(x^2\right)}{-\log(2)x^4+O\left(x^3\right)}\\\\
&=-\frac{2}{\log(2)}+O\left(\frac1x\right)\\\\
&\to -\frac{2}{\log(2)}\,\,\text{( as}\,\,x\to \infty\,\text{)}
\end{align}$$
