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\title[]{MA140-Engineering Calculus}
\author[ Lecture 4 ]{\large{ Lecture 4 } 
\and %\\ \small{Supervisor: Dr. John Burns}
\\ 
 \includegraphics[scale=0.0]{Figures/lec14.jpg}}
%\date[June 2-4, 2016]{June 2-4, 2016}
%\institute[Uppsala University]{ }


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


\section{MA140}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%



\subsection{Limit}
\begin{frame}
\begin{example}
What is the difference between these two functions:
$$\frac{x^{2}-1}{x-1} \quad and \quad x+1$$
\end{example}
{\textcolor{blue}{we simplify $\frac{x^{2}-1}{x-1}$ $$\frac{x^{2}-1}{x-1}=\frac{(x-1)(x+1)}{x-1}$$ now the functions seem to be equal, the only difference between them is that tha rational function is undefined at $x=1$, so $$\frac{x^{2}-1}{x-1}=x+1 \quad for \quad x\neq 1$$    }}
\end{frame}
\begin{frame}
\begin{example}
How does the function $$f(x)=\frac{x^{2}-1}{x-1} $$ behave near $x=1$
\end{example}
{\textcolor{blue}{We have seen that the function $f$ equals $x+1$ except at $x=1$, so the graph of $f$ is thus the line $x+1$ with the point $(1,2)$ removed.}}
\begin{center}
			\includegraphics[scale=0.2]{Figures/lec27.pdf}
		\end{center}
\end{frame}
\begin{frame}
\begin{center}
			\includegraphics[scale=0.3]{Figures/lec29.pdf}
		\end{center}
		$$\lim_{x \to 1} f(x)=\lim_{x \to 1} \frac{x^{2}-1}{x-1}=2$$
\end{frame}
\begin{frame}
\begin{center}
			\includegraphics[scale=0.4]{Figures/lec30.pdf}
		\end{center}
\end{frame}
\begin{frame}{Finding the Limits}
Sometimes $\lim_{x \to a}f(x)$ can be evaluated by calculating $f(a)$. This holds, for example,
whenever $f(x)$ is an algebraic combination of polynomials and trigonometric functions
for which $f(a)$ is defined.
\begin{itemize}
\item $\lim_{x \to 3} x=3$ 
\item $\lim_{x \to 3} 5=5$ 
\item $\lim_{x \to 3} (5x-2)=13$
\end{itemize}
{\textbf {\textcolor{red}{Note:}}} The Identity and Constant Functions Have Limits at Every Point.
\end{frame}
\begin{frame}{The Limit Laws}
 If $L, M, a $ and $k$ are real numbers and $$\lim_{x \to a} f(x)=L \quad and \quad  \lim_{x \to a} g(x)=M$$
then:
\begin{itemize}
\item \textit{Sum Rule}: The limit of the sum of two functions is the sum of their limits.{\textcolor{blue}{$$\lim_{x \to a} [f(x)+g(x)]=\lim_{x \to a} f(x)+\lim_{x \to a} g(x)=L+M$$}}
\item \textit{Difference Rule:} The limit of the difference of two functions is the difference of their limits. {\textcolor{blue}{$$\lim_{x \to a} [f(x)-g(x)]=\lim_{x \to a} f(x)-\lim_{x \to a} g(x)=L-M$$}}
\end{itemize}
\end{frame}
\begin{frame}
\begin{itemize}
\item \textit{Product Rule:} The limit of a product of two functions is the product of their limits. {\textcolor{blue}{$$\lim_{x \to a} [f(x) \cdot g(x)]=(\lim_{x \to a} f(x))\cdot (\lim_{x \to a} g(x))=L\cdot M$$}}
\item \textit{Constant Multiple Rule:} The limit of a constant times a function is the constant times the limit of the function. {\textcolor{blue}{$$\lim_{x \to a} (k\cdot f(x)) =k\cdot (\lim_{x \to a} f(x)=k\cdot L$$}}
\item \textit{Quotient Rule:} The limit of a quotient of two functions is the quotient of their limits, provided the limit of the denominator is not zero.{\textcolor{blue}{$$\lim_{x \to a} \frac{f(x)}{g(x)}=\frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}=\frac{L}{M}, \quad M\neq 0$$}}
\end{itemize}

\end{frame}
\begin{frame}
\begin{itemize}
\item \textit{Power Rule:} If $r$ and $s$ are integers with no common factor and $s\neq 0,$ then {\textcolor{blue}{$$\lim_{x \to a} (f(x))^{r/s}=(\lim_{x \to a} f(x))^{r/s}=L^{r/s}$$}} 
\end{itemize}
\begin{example}
\begin{itemize}
\item $\lim_{x \to 1} (x^{3}+4x^{2}-3)=\lim_{x \to 1}x^{3}+\lim_{x \to 1} 4x^{2}-\lim_{x \to 1}3=1+4-3=2$
\end{itemize}
\end{example}
\begin{example}
\begin{itemize}
\item $\displaystyle{\lim_{x \to 1} \frac{x^{4}+x^{2}-1}{x^{2}+5}=\frac{\lim_{x \to 1} (x^{4}+x^{2}-1) }{\lim_{x \to 1} (x^{2}+5)}=\frac{1}{6}}$
\end{itemize}
\end{example}
\begin{example}
\begin{itemize}
\item $\lim_{x \to 1} \sqrt{4x^{2}-3}=\sqrt{\lim_{x \to 1} (4x^{2}-3)}=1$
\end{itemize}
\end{example}

\end{frame}






















\end{document}