<s>[INST] Solve $\lim_{x\to\infty}(x-x^2 \ln{\frac{1+x}{x}})$ without L'Hopital I've seen the solution to $\lim_{x\to\infty}(x-x^2 \ln{\frac{1+x}{x}})$ using L'Hopital and I was wondering if there's a way to find out the result without it. My initial attempt was outright stupid of me because I tried to substitute the limit of $\frac{\ln{(1+\frac{1}{x})}}{\frac{1}{x}}$ as $x$ approaches $\infty$ with $1$, which results in the initial limit being $0$. That's obviously false as I ignored the fact that I cannot do such a substitution when the limit is in an indeterminate form. That being said, how could you solve this limit without L'Hopital?
 [/INST] Use substitution $e^t = \frac{x}{1 + x}$;
$$x - x^2\ln\left(\frac{1+x}{x}\right) \quad \overset{e^t = \frac{x}{1 + x}}{\longrightarrow} \quad \frac{1}{e^{-t} - 1} + \frac{t}{\left( e^{-t} - 1 \right)^2} = \frac{t + e^{-t} - 1}{\left( e^{-t} - 1 \right)^2} \\ \quad \\ \quad \\ = \frac{te^{t} + 1 - e^{t}}{e^{t}\left( e^{-t} - 1 \right)^2} \\ \quad \\ \quad \\ = \frac{te^{t} + 1 - e^{t}}{\left( e^{\frac{-t}{2}} - e^{\frac{t}{2}} \right)^2} \\ \quad \\ \quad \\ = \frac{te^{t} + 1 - e^{t}}{\left( e^{\frac{-t}{2}} + e^{\frac{t}{2}} \right)^2 - 4} \\ \quad \\ \quad \\ = \frac{te^{t} - e^{\frac{t}{2}}\left( e^{\frac{-t}{2}} + e^{\frac{t}{2}} - 2 \right) - 2e^{\frac{t}{2}} + 2}{\left( e^{\frac{-t}{2}} + e^{\frac{t}{2}} - 2 \right)\left( e^{\frac{-t}{2}} + e^{\frac{t}{2}} + 2 \right)} \\ \quad \\ \quad \\ = \frac{te^{t} - 2e^{\frac{t}{2}} + 2}{\left( e^{\frac{-t}{2}} + e^{\frac{t}{2}} - 2 \right)\left( e^{\frac{-t}{2}} + e^{\frac{t}{2}} + 2 \right)} - \frac{e^{\frac{t}{2}}}{\left( e^{\frac{-t}{2}} + e^{\frac{t}{2}} + 2 \right)}$$
As $x \to \infty, t \to 0$. WLOG assume $te^{\frac{t}{2}} + 1 \approx e^{t}$ for small $t$. (The intention behind this was to be able to factor the numerator while keeping the integrity of limit equivalent. Since, $\color{red}{x}e^{x}$ would not be factored easily, the goal was to approximate $\color{red}{x}e^{x} \approx \color{blue}{k}e^{\frac{x}{\color{blue}{c}}} + \color{blue}{p}$, where $k = 1$ was preferred, for small $t$).
Hence, from where we left off,
$$\frac{te^{t} - 2e^{\frac{t}{2}} + 2}{\left( e^{\frac{-t}{2}} + e^{\frac{t}{2}} - 2 \right)\left( e^{\frac{-t}{2}} + e^{\frac{t}{2}} + 2 \right)} - \frac{e^{\frac{t}{2}}}{\left( e^{\frac{-t}{2}} + e^{\frac{t}{2}} + 2 \right)} \\ \quad \\ \quad \\ = \frac{e^{\frac{t}{2}} \left( te^{\frac{t}{2}} + 1 \right) - 3e^{\frac{t}{2}} + 2}{\left( e^{\frac{-t}{2}} + e^{\frac{t}{2}} - 2 \right)\left( e^{\frac{-t}{2}} + e^{\frac{t}{2}} + 2 \right)} - \frac{e^{\frac{t}{2}}}{\left( e^{\frac{-t}{2}} + e^{\frac{t}{2}} + 2 \right)} \\ \quad \\ \quad \\ \approx \frac{e^{\frac{t}{2}}e^{t} - 3e^{\frac{t}{2}} + 2}{\left( e^{\frac{-t}{2}} + e^{\frac{t}{2}} - 2 \right)\left( e^{\frac{-t}{2}} + e^{\frac{t}{2}} + 2 \right)} - \frac{e^{\frac{t}{2}}}{\left( e^{\frac{-t}{2}} + e^{\frac{t}{2}} + 2 \right)} $$
Now, let $u = e^{\frac{t}{2}}$;
$$\frac{e^{\frac{t}{2}}e^{t} - 3e^{\frac{t}{2}} + 2}{\left( e^{\frac{-t}{2}} + e^{\frac{t}{2}} - 2 \right)\left( e^{\frac{-t}{2}} + e^{\frac{t}{2}} + 2 \right)} - \frac{e^{\frac{t}{2}}}{\left( e^{\frac{-t}{2}} + e^{\frac{t}{2}} + 2 \right)} \\ \quad \\ \quad \\ \overset{u = e^{\frac{t}{2}}}{\longrightarrow} \frac{u^3 - 3u + 2}{\left( u^{-1} + u + 2 \right)\left( u^{-1} + u - 2 \right)} - \frac{u}{\left( u^{-1} + u + 2 \right)} \\ \quad \\ \quad \\ = \frac{u^2 (u - 1)^2 (u + 2)}{\left( 1 + u^2 + 2u \right)\left( 1 + u^2 - 2u \right)} - \frac{u}{\left( u^{-1} + u + 2 \right)} \\ \quad \\ \quad \\ = \frac{u^2 (u - 1)^2 (u + 2)}{(u + 1)^2(u - 1)^2} - \frac{u^2}{(u + 1)^2} \\ \quad \\ \quad \\ = \frac{u^2(u + 1)}{(u + 1)^2} = \frac{u^2}{u + 1}$$
As $x \to \infty \Rightarrow t \to 0 \Rightarrow u \to 1$
Hence and finally, $$\bbox[5px,border:2px solid black]{\lim\limits_{u \to 1}{\frac{u^2}{u + 1}} = \frac{1}{2}}$$
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