Answer
a) $-21$
b) $49$
c) $0$
d) $1$
e) $1$
f) $7$
g) $-7$
h) $-1/7$
Work Step by Step
$\lim_{t\to t_0}f(t)=-7$ and $\lim_{t\to t_0}g(t)=0$
a) $\lim_{t\to t_0}3f(t)=3\lim_{t\to t_0}f(t)=3\times(-7)=-21$ (Constant Multiple Rule)
b) $\lim_{t\to t_0}(f(t))^2=(\lim_{t\to t_0}f(t))^2=(-7)^2=49$ (Power Rule)
c) $\lim_{t\to t_0}[f(t)\times g(t)]=\lim_{t\to t_0}f(t)\times\lim_{t\to t_0}g(t)=(-7)\times0=0$ (Product Rule)
d) $\lim_{t\to t_0}\frac{f(t)}{g(t)-7}=\frac{\lim_{t\to t_0}f(t)}{\lim_{t\to t_0}g(t)-\lim_{t\to t_0}7}=\frac{-7}{0-7}=\frac{-7}{-7}=1$ (Quotient and Difference Rule)
e) $\lim_{t\to t_0}\cos(g(t))=\cos(\lim_{t\to t_0}g(t))=\cos0=1$ (Composite Rule)
f) $\lim_{t\to t_0}|f(t)|=|\lim_{t\to t_0}f(t)|=|-7|=7$ (Composite Rule)
g) $\lim_{t\to t_0}(f(t)+g(t))=\lim_{t\to t_0}f(t)+\lim_{t\to t_0}g(t)=-7+0=-7$ (Sum Rule)
h) $\lim_{t\to t_0}\frac{1}{f(t)}=\frac{1}{\lim_{t\to t_0}f(t)}=\frac{1}{-7}=-\frac{1}{7}$ (Quotient Rule)
