3
$\begingroup$

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.

$\endgroup$
10
$\begingroup$

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 $h\to0$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.