Proof :

We have,

lim\(_{θ\to 0} { \dfrac {\mathrm tan \mathrm θ}{ \mathrm θ} }   \) = lim\(_{θ\to 0} { \dfrac {\mathrm \sin \mathrm θ} {\mathrm θ \mathrm \cos\mathrm θ} }   \)      \( \{∵ \tan\theta =  \dfrac {\sin\theta}{\cos\theta}  \} \)

 = lim\(_ \mathrm {θ\to 0}  \dfrac {\mathrm{\sin θ} } { \mathrm θ} \) × lim\(_ \mathrm {θ\to 0} \mathrm{cos θ} \)    \( \{ ∵\) lim\(_{x\to y}f(x)g(x)\) = lim\(_{x\to y}f(x)\)  . lim\(_{x\to y}g(x) \} \) 

= 1 × 1

= 1

Hence,

lim\(_{θ\to 0} \tan\)θ = 1