Calculate $\;\lim\limits_{x\to\infty} \left[x^2\left(1+\dfrac1x\right)^x-ex^3\ln\left(1+\dfrac1x\right)\right]$.

It is a $\frac00$ case of indetermination if we rewrite as $\lim_{x\to\infty} \frac{((1+\frac1x)^x-e\ln(1+\frac1x)^x)}{\frac{1}{x^2}}$, since $\lim_{x\to\infty} \frac{1}{x^2} = 0$, $\lim_{x\to\infty} (1+\frac1x)^x = e$ and $\lim_{x\to\infty} \ln(1+\frac1x)^x = 1$.

I think that it is the type that has a solution without l'Hospital's rule, but it's quite difficult to find, so l'Hospital still remains the best try to me. I tried using it with different rewrites, but it seems that it needs to be used multiple times, and the expression gets harder and harder to calculate, so I assume that some other limit must be applied first to make the expression nicer.

Also, I futilely tried to use the following known limits by changing $x$ into $y = \frac1x$ if needed (and adding and substracting $ex^2$ in the main parenthesis and trying to use the last 2 limits), but maybe it can help you: $\lim_{x\to0} \frac{a^x-1}{x} = \ln a$, $\lim_{x\to0} \frac{\ln(1+x)}{x} = 1$, $\lim_{x\to0} \frac{(1+x)^r-1}{x} = r$, $\lim_{x\to0} \frac{(1+x)^\frac1x-e}{x} = -\frac{e}{2}$, $\lim_{x\to\infty} (x-x^2ln(1+\frac1x)) = \frac12$.

Can you help me with this problem?

  • $\begingroup$ Are you allowed to use Taylor series? $\endgroup$
    – angryavian
    Dec 9, 2020 at 18:37
  • $\begingroup$ Yes! I would prefer a solution without it because I'm not very familiar with Taylor series, but I'm grateful for anything you have. I can learn more about it later :) $\endgroup$ Dec 9, 2020 at 18:41

2 Answers 2


I am going to calculate your limit without using Taylor series but only the two following notable limits:





Now I am going to calculate the first limit by using the following substitution:



$=\lim\limits_{t\to0}\dfrac{e(1+t)-e\ln\big[e(1+t)\big] }{\left\{1-\ln\big[e(1+t)\big]\right\}^2}=$




$=e\cdot\dfrac12\cdot1^2=\dfrac e2\;.$

Now I am going to calculate the second limit by using the following substitution:









$=\dfrac e2\cdot\dfrac14=\dfrac e8\;.$

  • $\begingroup$ Wow! Thank you very much! It is a very smart and elementary solution! I really appreciate your effort of coming up with such a solution! Really awesome! $\endgroup$ Dec 9, 2020 at 20:55
  • $\begingroup$ You are welcome. $\endgroup$
    – Angelo
    Dec 9, 2020 at 20:57

We can use the following asymptotic relations

$$\left(1+\frac{1}{x}\right)^x = e - \frac{e}{2x}+\frac{11e}{24x^2}+O\left(\frac{1}{x^3}\right)$$

$$\ln\left(1+\frac{1}{x}\right) = \frac{1}{x}-\frac{1}{2x^2}+\frac{1}{3x^3}+O\left(\frac{1}{x^4}\right)$$

to get that

$$x^2\left(1+\frac{1}{x}\right)^x-ex^3\ln\left(1+\frac{1}{x}\right) \sim x^2\left(e - \frac{e}{2x}+\frac{11e}{24x^2} - e+\frac{e}{2x}-\frac{e}{3x^2}\right) = \frac{e}{8}$$

  • $\begingroup$ Thank you very much for your solution! As I mentioned earlier, I don't know much about Taylor series, but I appreciate your solution very much! I will learn more about it as soon as possible seeing how useful it is and how much it shortens the solution! $\endgroup$ Dec 9, 2020 at 18:45
  • $\begingroup$ @MichaelGoldberg I think what can point you in that direction is knowing which limits to look for. The log Taylor series is derived from geometric series, but for the other limit you have to find the equivalent of $\lim_{x\to 0}\frac{(1+x)^{\frac{1}{x}}-e}{x}$ but for the second order term, something like $\lim_{x\to 0}\frac{x(1+x)^{\frac{1}{x}}-ex+\frac{e}{2}}{x^2}$ $\endgroup$ Dec 9, 2020 at 18:47
  • $\begingroup$ Thanks a lot! I will try to use this approach! $\endgroup$ Dec 9, 2020 at 18:59

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