Solución

lim⁡θ→πsenθtgθ=senπtgπ=00\mathop {\lim }\limits_{\theta \to \pi } {{sen\theta } \over {tg\theta }} = {{sen\pi } \over {tg\pi }} = {0 \over 0}

Así, lim⁡θ→πsenθtgθ=lim⁡θ→πsenθsenθcos⁡θ=lim⁡θ→πcos⁡θ=−1\mathop {\lim }\limits_{\theta \to \pi } {{sen\theta } \over {tg\theta }} = \mathop {\lim }\limits_{\theta \to \pi } {{sen\theta } \over {{{sen\theta } \over {\cos \theta }}}} = \mathop {\lim }\limits_{\theta \to \pi } \cos \theta = - 1