Question

1. calculate the following limits using continuity.

a. $lim_{\theta
ightarrowpi}\frac{cos(\theta)}{\theta + 1}$
b. $lim_{t
ightarrow1}(arctan(t)-(t + 1)^4)$

Explanation:

Step1: Recall continuity property

If a function $f(x)$ is continuous at $x = a$, then $\lim_{x
ightarrow a}f(x)=f(a)$.

Step2: Solve part A

For $\lim_{\theta
ightarrow\pi}\frac{\cos(\theta)}{\theta + 1}$, since the function $f(\theta)=\frac{\cos(\theta)}{\theta + 1}$ is continuous at $\theta=\pi$. Substitute $\theta=\pi$ into the function: $f(\pi)=\frac{\cos(\pi)}{\pi + 1}$. We know that $\cos(\pi)=- 1$, so $f(\pi)=\frac{-1}{\pi + 1}=-\frac{1}{\pi + 1}$.

Step3: Solve part B

For $\lim_{t
ightarrow1}(\arctan(t)-(t + 1)^4)$, since the functions $y_1=\arctan(t)$ and $y_2=(t + 1)^4$ are continuous at $t = 1$. Substitute $t = 1$ into the function. We know that $\arctan(1)=\frac{\pi}{4}$ and $(1 + 1)^4=16$. Then $\lim_{t
ightarrow1}(\arctan(t)-(t + 1)^4)=\frac{\pi}{4}-16$.

Answer:

A. $-\frac{1}{\pi + 1}$
B. $\frac{\pi}{4}-16$