Answer
$\lim\limits_{x \to \infty} f(x)=\dfrac{1}{2}$
$\lim\limits_{x \to -\infty} f(x)=-\dfrac{1}{2}$
Work Step by Step
We are given the function:
$f(x)=\dfrac{\sqrt{x^2+1}}{2x+1}$
Determine $\lim\limits_{x \to \infty} f(x)$:
$\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{\sqrt{x^2+1}}{2x+1}$
$=\lim\limits_{x \to \infty}\dfrac{\dfrac{\sqrt{x^2+1}}{x}}{\dfrac{2x}{x}+\dfrac{1}{x}}$
$=\lim\limits_{x \to \infty}\dfrac{\sqrt{\dfrac{x^2}{x^2}+\dfrac{1}{x^2}}}{2+\dfrac{1}{x}}$
$=\lim\limits_{x \to \infty}\dfrac{\sqrt{1+\dfrac{1}{x^2}}}{2+\dfrac{1}{x}}$
$=\dfrac{1}{2}$
Determine $\lim\limits_{x \to -\infty} f(x)$:
$\lim\limits_{x \to -\infty} f(x)=\lim\limits_{x \to -\infty} \dfrac{\sqrt{x^2+1}}{2x+1}$
$=\lim\limits_{x \to -\infty}\dfrac{\dfrac{\sqrt{x^2+1}}{x}}{\dfrac{2x}{x}+\dfrac{1}{x}}$
$=\lim\limits_{x \to -\infty}\dfrac{-\sqrt{\dfrac{x^2}{x^2}+\dfrac{1}{x^2}}}{2+\dfrac{1}{x}}$
$=\lim\limits_{x \to -\infty}\dfrac{-\sqrt{1+\dfrac{1}{x^2}}}{2+\dfrac{1}{x}}$
$=-\dfrac{1}{2}$
