Given: $\mathrm{f}(\mathrm{t})=\frac{\sin \mathrm{t}}{\mathrm{t}}$
At $t=0$, we will check continuity of the function.
$\begin{array}{l}
\mathrm{LHL}=\mathrm{f}(0-\mathrm{h}) \\
=\lim _{\mathrm{h} \rightarrow 0} \frac{\sin (0-\mathrm{h})}{(0-\mathrm{h})}=\lim _{\mathrm{h} \rightarrow 0} \frac{-\sin \mathrm{h}}{-\mathrm{h}}=1 \\
\mathrm{RHL}=\mathrm{f}(0+\mathrm{h}) \\
\lim _{\mathrm{h} \rightarrow 0} \frac{\sin (0+\mathrm{h})}{(0+\mathrm{h})} \\
=\lim _{\mathrm{h} \rightarrow 0} \frac{\sin \mathrm{h}}{\mathrm{h}}=1
\end{array}$
and $\mathrm{f}(0)=1$
$\mathrm{LHL}=\mathrm{RHL}=\mathrm{f}(0)$
So, the function is continuous at $\mathrm{t}=0$
Now, we check the function is maximum or minimum.
$\begin{array}{l}
\mathrm{f}^{\prime}(\mathrm{t})=\frac{1}{\mathrm{t}} \cos \mathrm{t}-\frac{1}{\mathrm{t}^{2}} \sin \mathrm{t} \\
\text { and } \mathrm{f}^{\prime \prime}(\mathrm{t})=\frac{-1}{\mathrm{t}} \sin \mathrm{t}-\frac{1}{\mathrm{t}^{2}} \cos \mathrm{t}-\frac{1}{\mathrm{t}^{2}} \cos \mathrm{t}+\frac{2}{\mathrm{t}^{3}} \sin \mathrm{t} \\
=\frac{-\sin \mathrm{t}}{\mathrm{t}}-\frac{2 \cos \mathrm{t}}{\mathrm{t}^{2}}+\frac{2 \sin \mathrm{t}}{\mathrm{t}^{3}}
\end{array}$
For maximum or minimum value of $\mathrm{f}(\mathrm{x})$, $\begin{aligned} & \text { put } \\ & \mathrm{f}^{\prime}(\mathrm{x})=0 \\ \Rightarrow & \frac{\cos \mathrm{t}}{\mathrm{t}}-\frac{\sin \mathrm{t}}{\mathrm{t}^{2}}=0 \Rightarrow \frac{\tan \mathrm{t}}{\mathrm{t}}=1 \end{aligned}$
Now $\lim _{t \rightarrow 0} f^{\prime \prime}(t)$
$\begin{array}{l}
=-\lim _{t \rightarrow 0}\left(\frac{\sin t}{t}\right)-2 \lim _{t \rightarrow 0}\left(\frac{t \cos t-\sin t}{t^{3}}\right) \\
=-1-2 \lim _{t \rightarrow 0}\left(\frac{\cos t-t \sin t-\cos t}{3 t^{2}}\right)
\end{array}$
[using L'Hospital rule]
$\begin{array}{l}
=-1+\frac{2}{3} \lim _{t \rightarrow 0} \frac{\sin t}{t} \\
=-1+\frac{2}{3} \times 1=\frac{-1}{3} < 0
\end{array}$
So, function $\mathrm{f}(\mathrm{t})$ is maximum at $\mathrm{t}=0$