Answer
$\lim\limits_{t \to \infty}(\sqrt{25t^2+2}-5t)=0$
Work Step by Step
$\lim\limits_{t \to \infty}(\sqrt{25t^2+2}-5t)=\lim\limits_{t \to \infty}(\sqrt{25t^2+2}-5t)\times \frac{\sqrt{25t^2+2}+5t}{\sqrt{25t^2+2}+5t}$
$=\lim\limits_{t \to \infty}\frac{25t^2+2-25t^2}{\sqrt{25t^2+2}+5t}$
$=\lim\limits_{t \to \infty}\frac{2}{\sqrt{25t^2+2}+5t}$
$=\lim\limits_{t \to \infty}\frac{2}{\sqrt{25t^2+2}+5t}\times \frac{1/t}{1/t}$
$=\lim\limits_{t \to \infty}\frac{2/t}{\sqrt{25+2/t^2}+5}$ (Use the properties of limits)
$=\frac{\lim\limits_{t \to \infty}2/t}{\sqrt{\lim\limits_{t \to \infty}(25+2/t^2)}+\lim\limits_{t \to \infty}5}$ (Use the limit $\lim\limits_{t \to \infty}k/t^n=0$ for $n=1,2,...$)
$=\frac{0}{\sqrt{25+0}+5}$
$=0$
Thus, $\lim\limits_{t \to \infty}(\sqrt{25t^2+2}-5t)=0$.
