Question

find the limit.
lim_{\theta
ightarrow0}\frac{cos(\theta)-1}{sin(\theta)}
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1. -/1 points

find the limit.
lim_{\theta
ightarrow0}\frac{sin(\theta)}{\theta + \tan(\theta)}

Explanation:

Step1: Use L - H rule (indeterminate form 0/0)

When $\theta
ightarrow0$, $\cos(\theta)-1
ightarrow0$ and $\sin(\theta)
ightarrow0$. The limit $\lim_{\theta
ightarrow0}\frac{\cos(\theta)-1}{\sin(\theta)}$ is in the $\frac{0}{0}$ form. By L - H rule, if $\lim_{x
ightarrow a}\frac{f(x)}{g(x)}$ is in $\frac{0}{0}$ or $\frac{\infty}{\infty}$ form, then $\lim_{x
ightarrow a}\frac{f(x)}{g(x)}=\lim_{x
ightarrow a}\frac{f'(x)}{g'(x)}$. Differentiate $f(\theta)=\cos(\theta)-1$ and $g(\theta)=\sin(\theta)$. $f'(\theta)=-\sin(\theta)$ and $g'(\theta)=\cos(\theta)$. So, $\lim_{\theta
ightarrow0}\frac{\cos(\theta)-1}{\sin(\theta)}=\lim_{\theta
ightarrow0}\frac{-\sin(\theta)}{\cos(\theta)}$.

Step2: Evaluate the new limit

Substitute $\theta = 0$ into $\frac{-\sin(\theta)}{\cos(\theta)}$. We get $\frac{-\sin(0)}{\cos(0)}=\frac{0}{1}=0$.

Step3: For the second limit

When $\theta
ightarrow0$, $\sin(\theta)
ightarrow0$ and $\theta+\tan(\theta)
ightarrow0$. The limit $\lim_{\theta
ightarrow0}\frac{\sin(\theta)}{\theta+\tan(\theta)}$ is in the $\frac{0}{0}$ form. Differentiate $f(\theta)=\sin(\theta)$ and $g(\theta)=\theta+\tan(\theta)$. $f'(\theta)=\cos(\theta)$ and $g'(\theta)=1 + \sec^{2}(\theta)$.

Step4: Evaluate the new - formed limit

Substitute $\theta = 0$ into $\frac{\cos(\theta)}{1+\sec^{2}(\theta)}$. Since $\cos(0) = 1$ and $\sec(0)=1$, we have $\frac{\cos(0)}{1+\sec^{2}(0)}=\frac{1}{1 + 1}=\frac{1}{2}$.

Answer:

The first limit $\lim_{\theta
ightarrow0}\frac{\cos(\theta)-1}{\sin(\theta)} = 0$
The second limit $\lim_{\theta
ightarrow0}\frac{\sin(\theta)}{\theta+\tan(\theta)}=\frac{1}{2}$