Problem (cf. exercise in L'Hospital section in Stewart's Calculus Book) is $$\lim_{x\rightarrow \infty} \bigg[x - x^2\ln\ \bigg(\frac{1+x}{x}\bigg)\bigg]$$

Note that $$ \lim_{x\rightarrow \infty} \ln\ \frac{e^x}{\bigg( \frac{1+x}{x} \bigg)^{x^2}} $$

Since $\lim_{x\rightarrow \infty} \bigg( 1+\frac{1}{x}\bigg)^x=e$, we can use L'Hospital : $$ \lim_{x\rightarrow \infty} \frac{e^x}{g(x)} =\lim_{x\rightarrow \infty} \frac{e^x}{g'(x)},\ g(x) =\bigg(\frac{1+x}{x}\bigg)^{x^2} $$

So by $\ln\ g = x^2\ln\ \bigg(1+\frac{1}{x}\bigg)$ and implicit differentiation, we can know by routine computation : $\lim_{x\rightarrow \infty} \frac{g^{(n)}(x)}{g(x)} = 1$. Hence we cannot find limit of original question.

But we can find an answer by Taylor : $$ \ln\ (1+t) = t-\frac{t^2}{2} + \frac{t^3}{3}-\cdots $$ So the answer is $\frac{1}{2}$. But by not using Taylor, can we find limit ? Thank you in advance.


Setting $x=\dfrac1h$

the limit becomes $$\lim_{h\to0}\frac{h-\ln(1+h)}{h^2}$$ which is of the form $\dfrac00$

So applying L'Hospital's $$\lim_{h\to0}\frac{1-\dfrac1{1+h}}{2h}=\lim_{h\to0}\dfrac h{2h(1+h)}$$

Now we can safely cancel out $h$ as $h\ne0$ as $h\to0$

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