I have 
```
\documentclass[14pt]{beamer}
\usetheme{Warsaw}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{graphicx}
\usepackage{pgf,tikz,pgfplots}
\usepackage{hyperref}
\hypersetup{unicode,colorlinks=true,allcolors=blue,bookmarksnumbered}
\newtheorem{question}{Question}
%\setbeamercovered{transparent} 
%\setbeamertemplate{navigation symbols}{} 
%\logo{} 
%\institute{} 
\date{\today} 
%\subject{} 
\begin{document}

\begin{frame}
\titlepage
\end{frame}

\begin{frame}
\tableofcontents
\end{frame}

\section{Calculte the folowing limits}
\subsection{Review}
\begin{frame}
	\begin{question}
	Calculate
	\[ L= \displaystyle\lim_{x \to 3}\dfrac{2 x^2-7 x+3}{9-x^2}. \]
	\end{question}
\end{frame}

\begin{frame}
	\begin{question}
	Calculate 
		\[ L_1= \displaystyle\lim_{x \to 1^{+}}\dfrac{3-2x}{x-1}, \quad L_2= \displaystyle\lim_{x \to 1^{-}}\dfrac{3-2x}{x-1}. \]
	\end{question}
\end{frame}

\begin{frame}
	\begin{question}
	\[ f(x) = \begin{cases}
	x + 3, &\text{ if } x <-1,\\
	3  - 5x, &\text{ if } x >-1.\\
	\end{cases} \]
	Calculate 
		\[ L_1= \displaystyle\lim_{x \to -1^{+}}f(x),\,  L_2= \displaystyle\lim_{x \to -1^{-}}f(x),\,  L = \lim_{x \to -1}f(x). \]
	\end{question}
\end{frame}


\begin{frame}
	\begin{question}
	Let be given the fuction 
		\[ f(x) = \begin{cases}
		\dfrac{x^2 - 5x+6}{x^2 - 4}, &\text{ if } x \neq 2,\\
		3, &\text{ if } x= 2.\\
		\end{cases} \]
		Calculate $ \displaystyle\lim_{x \to 2}f(x). $
		\end{question}
\end{frame}


\begin{frame}
	\begin{question}
		
	Calculate $ \displaystyle\lim_{x \to -\infty}\dfrac{\sqrt{2x^2 - 3x}}{x+3}. $
	\end{question}
\end{frame}

\begin{frame}
	\begin{question}
		
	Calculate $ \displaystyle\lim_{x \to -\infty}(\sqrt{x^2 + 3x} + 2x + 1). $
	\end{question}
\end{frame}
\end{document}

```

How to add `\subsection{Review}` in the form `1.1 Review ` instead of a ball like this?

![image.png](/image?hash=68036968fd7cd17afcabdfdddff601d3db4a5ff4ab05eaad5eb545b759393fec)