Answer
$i+3j-\pi k$
Work Step by Step
Given: $\lim\limits_{t \to 1} (\dfrac{t^{2}-t}{t-1}i+\sqrt{t+8}j+\dfrac{\sin\pi t}{\ln t}\mathrm{k})$
$\lim\limits_{t \to 1} \dfrac{t^{2}-t}{t-1}=\lim\limits_{t \to 1} \dfrac{t(t-1)}{t-1}=\lim\limits_{t \to 1} (t) =1; \lim\limits_{t \to 1}
\sqrt{t+8}=3$
and $\lim\limits_{t \to 1} \dfrac{\sin\pi t}{\ln t}=\dfrac{0}{0}$ [Need to apply L'Hospital's Rule].
This gives: $\lim\limits_{t \to 1} \dfrac{\pi\cos\pi t}{\dfrac{1}{t}}=\dfrac{\pi(-1)}{1}=-\pi$
Hence, $\lim\limits_{t \to 1} (\dfrac{t^{2}-t}{t-1}i+\sqrt{t+8}j+\dfrac{\sin\pi t}{\ln t}\mathrm{k})=i+3j-\pi k$
