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