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,

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

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