Limit of function using L'Hôpital's rule: $\lim_{x \to \infty}{x-x^2\ln(1+\frac{1}{x})}$

Question

I have to calculate $\lim_{x \to \infty}{x-x^2\ln(1+\frac{1}{x})}$. I rewrote it as $\lim_{x \to \infty}{\frac{x-x^3\ln^2(1+\frac{1}{x})}{1 + x\ln(1+\frac{1} {x})}}$ and tried to apply L'Hôpital's rule but it didn't work. How to end this?

Answer 1

Make the substitution $t=\dfrac{1}{x}$. Then, $x\to\infty \implies t\to 0$

$$\lim_{x\to\infty}\left(x-x^2\ln\left(1+\frac{1}{x}\right)\right)\\ = \lim_{t\to 0}\left(\frac{1}{t}-\frac{1}{t^2}\ln(1+t)\right)\\= \lim_{t\to 0}\left(\frac{t-\ln(1+t)}{t^2}\right)$$

This comes out as $\frac{0}{0}$ on direct plugging of values, so it's ready for some L'Hopital bash.

$$\lim_{t\to 0}\left(\dfrac{1-\frac{1}{1+t}}{2t}\right)$$

Again, by direct plugging, we get $\frac{0}{0}$, so recursion of L'Hopital will work.

$$\lim_{t\to 0}\left(\dfrac{0+\frac{1}{(1+t)^2}}{2}\right)=\lim_{t\to 0}\left(\frac{1}{2(1+t)^2}\right)=\boxed{\frac{1}{2}}$$

Answer 2

The reason why L'Hôpital's rule didn't work is that you added on a determined-term into the limit, i.e. $$ 1 + x \ln \left( 1 + \frac{1}{x} \right) = 1 + \ln \left( \left( 1 + \frac{1}{x} \right)^x \right) \to 1 + \ln \left( e \right)= 2 $$ when $x \to \infty$. Another approach with the one from Prasun Biswas is using the Taylor series for $1/x$, for $x>1$: \begin{align} x - x^2 \ln \left( 1 + \frac{1}{x} \right) &= x - x^2 \left( \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{ k} \frac{1}{x^k} \right) = x - x^2 \left( \frac{1}{x} - \frac{1}{2x^2} + \sum_{k=3}^{\infty} \frac{(-1)^{k+1}}{k} \frac{1}{x^k} \right) \\ &= \frac{1}{2} + \sum_{k=3}^{\infty} \frac{(-1)^{k+1}}{k} \frac{1}{x^{k-2}} \end{align} Letting $x \to \infty$, one ends up the limit with $\frac{1}{2}$.