I don't know how to solve the following limit without using series expansion.

$$\lim_{x\to 0} \frac{(1+x)^x -1 -x^2}{x^3} $$

I have tried use L'Hopital's rule and finding bounds to use squeeze theorem but to no avail. Please give me hints on how to compute the limit, instead of posting full answers. Thanks in advance :)

Edit: I have just found an answer, though it is not elegant at all. It is done simply by applying L'Hopital's rule three times and there is not much to say about it. Still, other answers are welcome.

  • $\begingroup$ I didn't evaluate it by myself, but I guess you may try $(1+x)^x=e^{\ln (1+x)^x}$, and with the usage of theorem about limit of composition of continuous functions $\endgroup$ – Eric Oct 19 '16 at 15:16
  • $\begingroup$ @Eric but it one would still arrive at $\frac 00$ right? $\endgroup$ – lEm Oct 19 '16 at 15:23
  • $\begingroup$ well.. my way seems not work! $\endgroup$ – Eric Oct 19 '16 at 15:40

Consider the limit with $x=1/p$, $p$ integer, $$\lim_{p\to\infty}p^3\left(\left(1+\frac1p\right)^{1/p}-1-\frac1{p^2}\right).$$

Multiplying/dividing by the conjugate multinomial we get

$$\lim_{n\to\infty}\frac{p^3\left(1+\frac1p-\left(1+\frac1{p^2}\right)^p\right) } {\left(1+\frac1p\right)^{(p-1)/p}+\left(1+\frac1p\right)^{(p-2)/p}\left(1+\frac1{p^2}\right)+\cdots\left(1+\frac1{p^2}\right)^{(p-1)/p}}.$$

By the binomial theorem, the numerator is the finite sum

$$p^3\left(1+\frac1p-1-\frac p{p^2}-\frac{p(p-1)}{2p^4}-\frac{p(p-1)(p-2)}{3!p^6}-\cdots\frac1{p^{2p}}\right)\\ =-\frac{(p-1)}{2}-\frac{(p-1)(p-2)}{3!p^2}-\cdots\frac1{p^{2p-3}}$$ while the denominator has $p$ terms that are all between $1$ and $1+1/p$, hence is between $p$ and $p+1$.

In the limit, a single term remains, $$-\frac12.$$



You can use the generalized binomial formula, which works for any exponent.

$$(1+x)^x\approx 1+x\cdot x+\frac{x(x-1)}2x^2+\frac{x(x-1)(x-2)}{3!}x^3+\cdots\binom xn{x^n}+\cdots$$

The coefficient of $x^3$ is $-\dfrac12$.

  • 1
    $\begingroup$ Only a note: The OP wants to have a solution without using series expansion. :-) $\endgroup$ – user90369 Oct 19 '16 at 15:57
  • 1
    $\begingroup$ @user90369: mh, I was hoping that without referring to Taylor nobody would notice... :-) $\endgroup$ – Yves Daoust Oct 19 '16 at 15:59
  • $\begingroup$ @YvesDaoust Right right... but it can be recognized as Taylor from a mile away ;-) $\endgroup$ – Skeleton Bow Oct 19 '16 at 16:02
  • $\begingroup$ @SkeletonBow: I have added a variant that only relies on the ordinary binomial theorem. $\endgroup$ – Yves Daoust Oct 19 '16 at 16:23
  • $\begingroup$ Ah yes, that's quite smart :) $\endgroup$ – Skeleton Bow Oct 19 '16 at 16:36

$x\to 0$ and $h\to 0$

$\displaystyle \frac{a^h-1}{h}\to \ln a$ , $\enspace$ the base $a$ is no critical point for $a>-1$

$\displaystyle \frac{(1+x)^x-1}{x^2}\to \frac{(1+x)^h-1}{hx}\to \frac{\ln (1+x)}{x}$

$\displaystyle \frac{(1+x)^x-1}{x^3}-\frac{1}{x}=\frac{\frac{(1+x)^x-1}{x^2}-1}{x}\to \frac{\frac{\ln (1+x)}{x}-1}{x}\to\frac{-x/2}{x}= -\frac{1}{2}$

$\displaystyle \frac{ (1+x)^x -1 -x^2}{x^3}\to -\frac{1}{2}+\frac{1}{x}-\frac{x^2}{x^3}=-\frac{1}{2}$

  • $\begingroup$ Thanks for the answer. However I am not entirely sure if one of the step is rigorous enough, as my class emphasizes heavily on being rigorous. In the answer, you basically takes a portion of the original function and takes the limit to makes it into another function, then takes the limit again. I am not entirely sure the validity3. $\endgroup$ – lEm Oct 19 '16 at 15:43
  • $\begingroup$ @Bubububu : You only need $\frac{a^x-1}{x}\to \ln a$. For the rest you can use L'Hopital's rule if you think that my steps are not clear. :-) $\endgroup$ – user90369 Oct 19 '16 at 15:50
  • $\begingroup$ "you basically takes a portion of the original function and takes the limit to makes it into another function, then takes the limit again" That's acceptable if the functions are continuous and inner function converges. $\endgroup$ – fleablood Oct 19 '16 at 16:40
  • $\begingroup$ By what logic are you getting that $\frac{\ln(1+x)}{x}-1 \to \frac{-x/2}{x}$?? I feel like there's a step missing there, or you need to justify that at least. The rest of the argument seems a little convoluted in the order it's in, but is definitely correct (in a very non-rigourous way) $\endgroup$ – Brevan Ellefsen Oct 19 '16 at 18:25
  • $\begingroup$ This may be intuitive and give a result, but in my opinion it is WRONG. You can say that $\frac{a^x-1}x \to \ln a$ implies $\dfrac{(1+y)^x-1}{xy} \to \dfrac{\ln(1+y)}y$ for any $y>-1$, but you can't take $x=y$, make both tend to zero and assume that the result is going to be the same. Remember that the function with value $0$ at the origin and $\dfrac{xy}{x^2+y^2}$ anywhere else tends to zero when $x\to 0$ for any value of $y$, but is constantly $\frac 12$ on $x=y$. As much as you can't do anything with $x$ and $y$, you can't take a limit with $x\to 0$ and still have $x$ on the result. $\endgroup$ – Emilio Oct 19 '16 at 20:27

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