Let $f ( x ) = \frac { 2 x } { \sqrt { x ^ { 2 } + 1 } }$ The set of all horizontal asymptotes of ƒ is

A) $\{ y = 2 \}$
B) $\{ y = - 2 \}$
C) $\{ y = 2 , y = - 2 \}$
D) $\{ y = - 1 , y = 1 \}$
E) $\varnothing$

Question

Let $f ( x ) = \frac { 2 x } { \sqrt { x ^ { 2 } + 1 } }$ The set of all horizontal asymptotes of ƒ is

A) $\{ y = 2 \}$
B) $\{ y = - 2 \}$
C) $\{ y = 2 , y = - 2 \}$
D) $\{ y = - 1 , y = 1 \}$
E) $\varnothing$

Quizplus · Accepted Answer

To find the horizontal asymptotes of the function $f(x) = \frac{2x}{\sqrt{x^2 + 1}}$, we examine the behavior of $f(x)$ as $x$ approaches infinity ($\infty$) and negative infinity ($-\infty$). As $x \to \infty$, the expression $\frac{2x}{\sqrt{x^2 + 1}}$ approaches $\frac{2x}{\sqrt{x^2}} = 2$, because the $x^2$ in the numerator and denominator dominate the behavior, making the square root of $x^2 + 1$ approximately equal to $\sqrt{x^2} = x$. Similarly, as $x \to -\infty$, the expression approaches $\frac{2(-x)}{\sqrt{(-x)^2}} = -2$, for the same reason. Therefore, the set of all horizontal asymptotes of $f$ is $\{y = 2, y = -2\}$.

Let f(x)=2xx2+1f ( x ) = \frac { 2 x } { \sqrt { x ^ { 2 } + 1 } }f(x)=x2+1​2x​

Let $f ( x ) = \frac { 2 x } { \sqrt { x ^ { 2 } + 1 } }$