Solve $\lim_{x\to\infty}(x-x^2 \ln{\frac{1+x}{x}})$ without L'Hopital

Question

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?

Answer 1

Let $$ \lim_{x \to \infty} \bigg\{x - x^2 \, \ln\left(\frac{1+x}{x}\right)\bigg \} $$ Now substitute $ x = \frac{1}{h}$

$$ \lim_{h \to 0} \bigg\{\frac{1}{h} - \frac{1}{h^2}\, \ln\left(1+h\right)\bigg \} $$

Now using Taylor series expansion of $\ln(1+x)$

$$ \lim_{h \to 0} \bigg\{ \frac{1}{h} - \frac{1}{h^2} \left( h - \frac{h^2}{2} + \frac{h^3}{3} - \cdots \right) \bigg \} $$

Now first 2 terms will get cancelled and after applying limit,

$$ \lim_{h \to 0} \bigg\{ \frac{1}{h} - \frac{1}{h} + \frac{1}{2} - \frac{h}{3} + \frac{h^2}{4} - \cdots \bigg \} $$

$$ = \frac{1}{2} $$

Answer 2

The limit can be solved without using l'Hopital's rule and Taylor expansion, in fact: $$\lim_{x\to\infty}\left(x-x^2 \log{\frac{1+x}{x}}\right)=\lim_{t \to 0}\frac{t-\log(1+t)}{t^2}$$ And the last limit is $1/2$ and can be prooven without l'Hopital's rule and Taylor expansion.

Answer 3

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, $$\bbox[5px,border:2px solid black]{\lim\limits_{u \to 1}{\frac{u^2}{u + 1}} = \frac{1}{2}}$$