Degrees Limit?

We know that $\displaystyle 180°=\pi$ . From these we can conclude that: $180°=\pi \\ \Rightarrow 1°=\cfrac { \pi }{ 180 } \\ \Rightarrow x°=\cfrac { \pi }{ 180 } x$

Then $\lim _{ x\rightarrow 0 }{ \cfrac { \sin { x° } }{ x } } \\ \Rightarrow \lim _{ x\rightarrow 0 }{ \cfrac { \sin { \left( \cfrac { \pi }{ 180 } x \right) } }{ x } }$

Clearly as $\displaystyle x\rightarrow 0$ ; $\displaystyle \left( \cfrac { \pi }{ 180 } x \right) \rightarrow 0$

Now let $\displaystyle y=\cfrac { \pi }{ 180 } x$

Then $\cfrac { \pi }{ 180 } .\lim _{ y\rightarrow 0 }{ \cfrac { \sin { y } }{ y } } \Rightarrow \cfrac { \pi }{ 180 } \\ \therefore l=\cfrac { \pi }{ 180 } \\ \& \quad l\times 180=\pi$ Hence the answer.

Using L'Hopital's Rule

$\lim _{ x\rightarrow 0 }{ \frac { sinx }{ x } } =\lim _{ x\rightarrow 0 }{ \frac { cosx }{ 1 } =\frac { 1 }{ 1 } } =1$