cant find solution

[hidden mango]

Davidikov

[loud siren]

$\lim_{12}{3}$ this works

[hidden mango]

Davidikov

[potent nacelle]

$2^{\lim\displaylimits_{12} 3}$

[hidden mango]

lstenger

[potent nacelle]

Uh, that's not what I expected to happen haha. LaTeX's forcing us to go with the “not-ugly” version apparently 😅

$2^{\lim\limits_{12} 3}$

[hidden mango]

lstenger

[potent nacelle]

there you go (but it's ugly)

[potent nacelle]

and disregard this: \displaylimits is an amsmath command useful to define operators that accept limits, see sec. 7.2 of the amsmath manual,

,texdoc amsldoc

[hidden mango]

Documentation for amsldoc: http://texdoc.net/pkg/amsldoc

[loud siren]

$2^{\displaystyle\lim_{12}{3}}$

[hidden mango]

Davidikov

[loud siren]

that works too

but underset works always

$\lim_{n\rightarrow\infty}{\left(\frac{n-10}{n+1}\right)^{3n+1}}=\lim_{n\rightarrow\infty}{\left(1+\frac{-11}{n+1}\right)^{3n+1}}=\lim_{n\rightarrow\infty}{\left(\left(1+\frac{-11}{n+1}\right)^{\frac{n+1}{-11}}\right)^{(3n+1)\frac{-11}{n+1}}}=e^{-11\cdot\displaystyle\lim_{n\rightarrow\infty}{\frac{3n+1}{n+1}}}=e^{-11\cdot\displaystyle\lim_{n\rightarrow\infty}\frac{3+\frac{1}{n}}{1+\frac{1}{n}}}=e^{-11\cdot3}=\frac{1}{e^{33}}$

[hidden mango]

Davidikov

[loud siren]

insane stuff

[potent nacelle]

Considering the snippet you wrote above, I'd personally avoid using e alone to typeset the expression, and instead do the following

$\exp\biggl( -11 \lim_{ n \rightarrow \infty} \frac{ 3+n^{-1} }{ 1+n^{-1} } \biggr)$

[hidden mango]

lstenger

[loud siren]

thats true too i just prefer e bec its look cool lol

[gritty sail]

No it's used exactly how \limits or \nolimits are used

it just means \limits when in display style and \nolimits otherwise, ie. the default tex behaviour

[potent nacelle]

til \(no)limits are TeX primitives, thanks @gritty sail !