Limits used as exponents -- how to write under the limit symbol without using \displaymath?

Question

If one needs to write a limit as an exponent, one might have this dilemma: if you use \displaymath, "x to infinity" will be nicely printed under the lim symbol, but your exponent will be using the normal font and will appear very big on the page. If you take out the \displaymath instruction, the exponent will use the small font, but now the part "x to infinity" is not a subscript to the "lim" symbol anymore, it just follows it. Trying to use any font size instructions with \displaystyle, or actually inside the math mode, does not seem to work for me! Does anybody know any trick to get around this?

This is the horrible expression I am fighting with (might be easier to make my point this way):

$e^{\left(\, \displaystyle \lim_{x \,\rightarrow\, \infty} 
\frac{\, 2x \sin{\frac{1}{x}} \,}{ 1 \,-\, \sin{\frac{1}{x}}} 
\,\right) } \,;$

I can't get the last exponent to behave, because it contains the limit (it's the last exponent, the one for the number e). If I take out the \displaystyle, the limit gets messed up, as explained above.

Answer 1

I would suggest you to use the exp(...) notation instead of e^{}; you can force the below position for the limit using \limits; since you are writing this expresion as in-line math I would also suggest using 1/x instead of \frac{1}{x}:

\documentclass{article}
\usepackage{amsmath}

\begin{document}

$\exp\left(\lim\limits_{x\rightarrow\infty} \frac{\, 2x \sin(1/x) \,}{ 1 - \sin(1/x)}\right)$

\end{document}

Answer 2

Such a big expression should probably go as a displayed equation:

\[
e^{\left(\lim\limits_{x\rightarrow\infty} 
  \frac{2x \sin\frac{1}{x}}{1-\sin\frac{1}{x}}\right)};
\]

but the result will be bad anyway, because the exponent will be much bigger than the basis.

I suggest you

\[
\exp\left(\lim_{x\to\infty}
  \frac{2x \sin\frac{1}{x}}{1-\sin\frac{1}{x}}\right);
\]

which has not the problem.

You are using many manual spaces which shouldn't be used.

Answer 3

The other answers give advice on how to re-write your expression. The advice works well in this instance as you can use exp, but there may be cases in the future when this is not an option- such as if the problem used 3 instead of e

If you would like to keep with the idea of keeping the limit in the exponent, then you can use \scalebox from the graphicx package to make it a bit more manageable.

\documentclass{article}
\usepackage{amsmath}
\usepackage{graphicx}
\begin{document}

$e^{\scalebox{1}{$\left( \displaystyle \lim_{x \rightarrow \infty} \dfrac{ 2x \sin{\dfrac{1}{x}} }{ 1 - \sin{\dfrac{1}{x}}} \right)$ }}$

\vspace{0.5cm}

$e^{\scalebox{0.9}{$\left( \displaystyle \lim_{x \rightarrow \infty} \dfrac{ 2x \sin{\dfrac{1}{x}} }{ 1 - \sin{\dfrac{1}{x}}} \right)$ }}$

\vspace{0.5cm}

$e^{\scalebox{0.8}{$\left( \displaystyle \lim_{x \rightarrow \infty} \dfrac{ 2x \sin{\dfrac{1}{x}} }{ 1 - \sin{\dfrac{1}{x}}} \right)$ }}$

\vspace{0.5cm}

$e^{\scalebox{0.7}{$\left( \displaystyle \lim_{x \rightarrow \infty} \dfrac{ 2x \sin{\dfrac{1}{x}} }{ 1 - \sin{\dfrac{1}{x}}} \right)$ }}$

\vspace{0.5cm}

$e^{\scalebox{0.6}{$\left( \displaystyle \lim_{x \rightarrow \infty} \dfrac{ 2x \sin{\dfrac{1}{x}} }{ 1 - \sin{\dfrac{1}{x}}} \right)$ }}$

\vspace{0.5cm}

$e^{\scalebox{0.5}{$\left( \displaystyle \lim_{x \rightarrow \infty} \dfrac{ 2x \sin{\dfrac{1}{x}} }{ 1 - \sin{\dfrac{1}{x}}} \right)$ }}$

\end{document}