At $t=0$, the function $f(t)=\frac{\sin t}{t}$ has

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 $\mathrm{t}=0$