Answer
$$\lim _{t \rightarrow 9} \frac{2 t-18}{5 t-45} =\frac{2}{5} $$
Work Step by Step
Given $$\lim _{t \rightarrow 9} \frac{2 t-18}{5 t-45}$$
let $$ f(t) = \frac{2 t-18}{5 t-45}$$
Since, we have
$$ f(9)=\frac{18-18}{45-45}=\frac{0}{0}$$
So, transform algebraically and cancel
\begin{aligned}
L&=\lim _{t \rightarrow 9} \frac{2 t-18}{5 t-45}\\
&=\lim _{t \rightarrow 9} \frac{2( t-9)}{5( t-9)}\\
&=\lim _{t \rightarrow 9} \frac{2}{5}\\
&=\frac{2}{5}
\end{aligned}
