Evaluate the limit limθ→0sin(θ2)θ\displaystyle{\lim\limits_{\theta \to 0} \dfrac{\sin{(\theta^2)}}{\theta}}θ→0limθsin(θ2) using the l'Hopital's Rule.
Since we have an indeterminate form of type 00\frac{0}{0}00, we can apply the l'Hopital's rule.
limθ→0sinθ2θ=limθ→0ddθ(sinθ2)ddθ(θ)=limθ→02θcosθ2=0Hence limθ→0sinθ2θ=0\lim_{\theta \to 0} \frac{\sin{\theta^2}}{\theta}=\lim_{\theta \to 0} \frac{\frac{d}{d\theta}(\sin{\theta^2})}{\frac{d}{d\theta}(\theta)}=\lim_{\theta \to 0}2 \theta cos{\theta^2}=0\\\\ \text{Hence }\\ \lim_{\theta \to 0} \frac{\sin{\theta^2}}{\theta}=0limθ→0θsinθ2=limθ→0dθd(θ)dθd(sinθ2)=limθ→02θcosθ2=0Hence limθ→0θsinθ2=0
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