You have to define yourself that command in your document preamble
\newcommand{\me}{\mathrm{e}}
Some comments
It's not necessary to declare this symbol as an ordinary symbol because by rule all math characters that are subject to alphabet changes are ordinary.
The construction \mathrm{e}
is essentially equivalent to
\begingroup\mathgroup=0 e\endgroup
because the Roman math alphabet corresponds to the math group 0. Each math alphabet (\mathrm
, \mathnormal
, \mathbf
and so on) corresponds to the choice of one among 16 math groups; \mathrm
is 0, \mathnormal
is 1, the others are from 4 onwards.
Each character has a \mathcode
which, for e
is "7165
which means
when an e
is found in math mode, it should be an ordinary symbol, taken from the font in math group 1 at slot "65
; but if \mathgroup
is set to a non negative value, use the character at slot "65
from the font in the specified math group (this because of the first digit "7
).
Thus calling \mathrm{e}
is like asking for a math code "0065
(numbers preceded by "
are hexadecimal): the first 0 means "ordinary symbol", the second one is the math group. With \mathbf{e}
we would ask for "0465
, according to the normal LaTeX setup.
Details about math codes are in the TeXbook or in TeX by Topic. Note that LaTeX calls \mathgroup
the primitive \fam
.
Best Answer
The macro for the limit operator is
\lim
.Without the
\
, it just treated as three charactersl
,i
,m
. This is no different that$xy$
representing a product of two termsx
, andy
, so$lim$
is a product of three terms:l
,i
,m
. So with$lim_{n\to\infty}$
, the subscript is applied to them
term. Perhaps the meaning is more obvious if you write and equivalent statement:Note that without the
\
the three letters are in italics, representing variables. The operator\lim
is not in italics representing an operator.