## Calculus: Early Transcendentals 8th Edition

$\lim\limits_{x\to\infty}\sqrt{x^2+1}=\infty$
$$\lim\limits_{x\to\infty}\sqrt{x^2+1}$$$$=\sqrt{\lim\limits_{x\to\infty}(x^2)+1}$$ As $x\to\infty$, $x^2\to\infty$. Therefore, $\lim\limits_{x\to\infty}(x^2)=\infty$ Which means, $\sqrt{\lim\limits_{x\to\infty}(x^2)+1}=\sqrt{\infty+1}=\infty$ In conclusion, $\lim\limits_{x\to\infty}\sqrt{x^2+1}=\infty$