4
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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
1
$\begingroup$

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, $$-\frac12.$$

$\endgroup$
3
$\begingroup$

Hint:

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 $-\dfrac12$.

$\endgroup$
5
  • 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

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.