Answer
$\frac{1}{\sqrt{2}}$
Work Step by Step
$\lim_{t\to\infty}\frac{t+3}{\sqrt{2t^2-1}}=\lim_{t\to\infty}\frac{t+3}{\sqrt{2t^2-1}}\times \frac{1/t}{1/t}$
$=\lim_{t\to\infty}\frac{1+3/t}{\sqrt{2-1/t^2}}$ (Use the properties of limits)
$=\frac{\lim_{t\to\infty}(1+3/t)}{\sqrt{\lim_{t\to\infty}(2-1/t^2)}}$ (Use the limit $=\lim_{t\to\infty} k/t=0$ and $\lim_{t\to\infty}k/t^2=0$)
$=\frac{1+0}{\sqrt{2-0}}$
$=\frac{1}{\sqrt{2}}$
Thus, $\lim_{t\to\infty}\frac{t+3}{\sqrt{2t^2-1}}=\frac{1}{\sqrt{2}}$
