2.1 Q-1

Q1. Use a graph to find the given limit, if it exists. (a) ${\lim }_{x\to 5}\sqrt{x-1}$

\section*{Solution:}

\begin{tabular}{|c|c|c|c|c|c|c|c|} \hline$x$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 \ \hline$y$ & 0 & 1 & 1.41 & 1.73 & 2 & 2.23 & 2.4 \ \hline \end{tabular}

From graph, it is clear that the limiting value of function approaches to 2 from both left and right side as $x\to 5$. i.e., ${\lim }_{x\to 5^{-}}f(x)=2={\lim }_{x\to 5^{+}}f(x)$ ⇒ Limit exists. Hence ${\lim }_{x\to 5}\sqrt{x-1}=2$ (b) ${\lim }_{x\to 1}\frac{x^2-1}{x-1}$

\section*{Solution:}

\begin{tabular}{|c|c|c|c|c|c|c|c|} \hline$x$ & -1 & -0.5 & 0 & 0.5 & 1 & 1.5 & 2 \ \hline$y$ & 0 & 0.5 & 1 & 1.5 & $\infty$ & 2.5 & 3 \ \hline \end{tabular}

From graph, it is clear that the limiting value of function approaches to 2 from both left and right sides as $x$ approaches to 1 . i.e., ${\lim }_{x\to 1^{-}}f(x)=2={\lim }_{x\to 1^{+}}f(x)$ ⇒ Limit exists. Hence ${\lim }_{x\to 1}\frac{x^2-1}{x-1}=2$ (c) ${\lim }_{x\to 0}\frac{x^2-3x}{x}$

\section*{Solution:}

\begin{tabular}{|c|c|c|c|c|c|c|c|} \hline$x$ & $\cdots$ & -1 & 0 & 1 & 2 & 3 & $\cdots$ \ \hline$y$ & $\cdots$ & -4 & \begin{tabular}{c} not \ defined \end{tabular} & -2 & 1 & 0 & $\cdots$ \ \hline \end{tabular}

From graph, it is clear that the limiting value of function approaches to -3 from both left and right sides as $x$ approaches to 0 . i.e., ${\lim }_{x\to 0^{-}}f(x)=-3={\lim }_{x\to 0^{+}}f(x)$ Hence ${\lim }_{x\to 0}\frac{x^2-3x}{x}=-3$ (d) ${\lim }_{x\to 0}\frac{|x|}{x}$

\section*{Solution:}

\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|} \hline$x$ & $\dots$ & -3 & -2 & -1 & 0 & 1 & 2 & 3 & $\dots$ \ \hline$y$ & $\dots$ & -1 & -1 & -1 & \begin{tabular}{c} not \ defined \end{tabular} & 1 & 1 & 1 & $\dots$ \ \hline \end{tabular}

From graph, it is clear that the limiting value of function approaches to -1 as $x$ approaches to 0 from left side. i.e., i.e., \begin{equation*} \lim _{x \rightarrow 0^{-}} f(x)=-1 \end{equation*}

Also the limiting value of function approaches to 1 as $x$ approaches to 0 from right side. i.e., \begin{equation*} \lim _{x \rightarrow 0^{+}} f(x)=1 \end{equation*}

From eq. (1) and (2), it is clear that ${\lim }_{x\to 0^{-}}f(x)\ne {\lim }_{x\to 0^{+}}f(x)$

So, limit does not exist. (e) ${\lim }_{x\to 2}f(x)$, where $f(x)=\{\begin{array}{cc}x& x<2\\ x+1& x\ge 2\end{array}$

\section*{Solution:}

Case-I: When $x<0$ then $f(x)=x$, so we have

\begin{tabular}{|c|c|c|c|c|c|} \hline$x$ & $\dots$ & -1 & 0 & 1 & +1.5 \ \hline$f(x)$ & $\dots$ & -1 & 0 & 1 & 1.5 \ \hline \end{tabular}

Case-II: When $x\ge 2$ then $f(x)=x+1$, so we have

\begin{tabular}{|c|c|c|c|c|} \hline$x$ & 2 & 3 & 4 & $\dots$ \ \hline$f(x)$ & 3 & 4 & 5 & $\dots$ \ \hline \end{tabular}

From graph, it is clear that the limiting value of function approaches to 2 as $x$ approaches to 2 from left side $(x<2)$. i.e., \begin{equation*} \lim _{x \rightarrow 2^{-}} f(x)=2 \end{equation*}

Also the limiting value of function approaches to 3 from right side as $x$ approaches to 2 . ( $x\ge 2$ ) i.e., \begin{equation*} \lim _{x \rightarrow 2+} f(x)=3 \end{equation*}

From eq. (1) and (2) ${\lim }_{x\to 2^{-}}f(x)\ne {\lim }_{x\to 2^{+}}f(x)$

So, limit does not exists. (f) ${\lim }_{x\to 0}f(x)$, where $f(x)=\{\begin{array}{cc}{x}^2& x<0\\ 2& x=0\\ \sqrt{x}-1& x>0\end{array}$

\section*{Solution:}

Case-I: When $x<0$ then $f(x)=x^2$

\begin{tabular}{|c|c|c|c|c|c|} \hline$x$ & $\dots$ & -3 & -2 & -1 & -0.5 \ \hline$f(x)$ & $\dots$ & 9 & 4 & 1 & 0.25 \ \hline \end{tabular}

Case-II: When $x>0$ then $f(x)=\sqrt{x}-1$

\begin{tabular}{|c|c|c|c|c|c|} \hline$x$ & 0.5 & 1 & 2 & 3 & $\dots$ \ \hline$f(x)$ & -0.29 & 0 & 0.41 & 0.73 & $\dots$ \ \hline \end{tabular}

From graph, it is clear that the limiting value of function approaches to 0 as $x$ approaches to 0 from left side ( $x<0$ ). i.e., \begin{equation*} \lim _{x \rightarrow 0^{-}} f(x)=0 \end{equation*}

Also the limiting value of function approaches to -1 from right side as $x$ approaches to $0.(x>0)$ i.e., \begin{equation*} \lim _{x \rightarrow 0^{+}} f(x)=-1 \end{equation*}

From eq. (1) and (2) ${\lim }_{x\to 0^{-}}f(x)\ne {\lim }_{x\to 0^{+}}f(x)$

So, limit does not exists. (g) ${\lim }_{x\to 0}\frac{1-\cos }{x^2}$

\section*{Solution:}

\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|} \hline$x$ & $\dots$ & -4 & -3 & -2 & -1 & 0 & 1 & 2 & 3 & 4 & $\dots$ \ \hline$y$ & $\dots$ & 0.1 & 0.22 & 0.35 & 0.46 & \begin{tabular}{c} not \ defined \end{tabular} & 0.46 & 0.35 & 0.46 & 0.1 & \ \hline \end{tabular}

From graph, it is clear that the limiting value of function approaches to $1/2$ from both left and right side as $x$ approaches to zero. i.e., ${\lim }_{x\to 0^{-}}\frac{1-\cos x}{x^2}=\frac{1}{2}={\lim }_{x\to 0^{+}}\frac{1-\cos x}{x^2}$ ⇒ Limit exists. Hence ${\lim }_{x\to 0}\frac{1-\cos x}{x^2}=\frac{1}{2}$