Answer
$$\frac{4}{3}$$
Work Step by Step
Given $$\lim _{t \rightarrow 1} \frac{t^{3}+t^{2}-5 t+3}{t^{3}-3 t+2}$$
Since $\lim _{t \rightarrow 1} \frac{t^{3}+t^{2}-5 t+3}{t^{3}-3 t+2}=\frac{0}{0}$, then
\begin{align*}
\lim _{t \rightarrow 1} \frac{t^{3}+t^{2}-5 t+3}{t^{3}-3 t+2}&=\lim _{t \rightarrow 1} \frac{(t-1)(t^2+2t-3)}{(t-1)(t^2+t-2)}\\
&=\lim _{t \rightarrow 1} \frac{ (t-1)(t+3)}{(t-1)(t+2)}\\
&=\lim _{t \rightarrow 1} \frac{ (t+3)}{ (t+2)}\\
&=\frac{4}{3}
\end{align*}
