Answer
$i+3j-\pi k$
Work Step by Step
Find the limit for $\lim\limits_{t \to 1} (\dfrac{t^{2}-t}{t-1}i+\sqrt{t+8}j+\dfrac{\sin\pi t}{\ln t}\mathrm{k})$
$\lim\limits_{t \to 1} \dfrac{t^{2}-t}{t-1} =1$ and $\lim\limits_{t \to 1}
\sqrt{t+8}=3$
Now, apply L'Hospital's Rule.
$\lim\limits_{t \to 1} \dfrac{\sin\pi t}{\ln t}=\dfrac{0}{0}$
$\lim\limits_{t \to 1} \dfrac{\pi\cos\pi t}{(1/t)}=-\dfrac{\pi \times (1)}{1}=-\pi$
Hence, we get $\lim\limits_{t \to 1} (\dfrac{t^{2}-t}{t-1}i+\sqrt{t+8}j+\dfrac{\sin\pi t}{\ln t}\mathrm{k})=i+3j-\pi k$
