Question

e $limlimits_{\theta \to 0}\frac{1 - cos\theta}{\theta}$.

Explanation:

Step1: Use L'Hopital's Rule

Since $\lim_{\theta\to0}\frac{1 - \cos\theta}{\theta}$ is in the $\frac{0}{0}$ form ($1-\cos0 = 0$ and $\theta = 0$ as $\theta\to0$), by L'Hopital's Rule, if $\lim_{x\to a}\frac{f(x)}{g(x)}$ is in $\frac{0}{0}$ or $\frac{\infty}{\infty}$ form, then $\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f^{\prime}(x)}{g^{\prime}(x)}$.
Let $f(\theta)=1-\cos\theta$, then $f^{\prime}(\theta)=\sin\theta$. Let $g(\theta)=\theta$, then $g^{\prime}(\theta) = 1$.
So, $\lim_{\theta\to0}\frac{1 - \cos\theta}{\theta}=\lim_{\theta\to0}\frac{\sin\theta}{1}$.

Step2: Evaluate the limit

Substitute $\theta = 0$ into $\frac{\sin\theta}{1}$. We know that $\sin0=0$.

Answer:

$0$