Calculate the following limits and identify the horizontal asymptotes (if any) for the function h ( x ) = 2 x 3 − 9 x 4 + 4 x 2 h\left(x\right)=\frac{2x^3-9}{x^4+4x^2}^{} h ( x ) = x 4 + 4 x 2 2 x 3 − 9 : lim ⁡ x → ∞ h ( x ) \lim_{x\rightarrow\infty}h\left(x\right) lim x → ∞ h ( x ) lim ⁡ x → − ∞ h ( x ) \lim_{x\rightarrow-\infty}h\left(x\right) lim x → − ∞ h ( x )

lim ⁡ x → ∞ h ( x ) \lim_{x\rightarrow\infty}h\left(x\right) lim x → ∞ h ( x ) = 0 =0 = 0 lim ⁡ x → − ∞ h ( x ) \lim_{x\rightarrow-\infty}h\left(x\right) lim x → − ∞ h ( x ) = 0 =0 = 0 Horizontal asymptote: y = 0 y=0 y = 0

lim ⁡ x → ∞ h ( x ) \lim_{x\rightarrow\infty}h\left(x\right) lim x → ∞ h ( x ) = 1 =1 = 1 lim ⁡ x → − ∞ h ( x ) \lim_{x\rightarrow-\infty}h\left(x\right) lim x → − ∞ h ( x ) = − 1 =-1 = − 1 No horizontal asymptote

lim ⁡ x → ∞ h ( x ) \lim_{x\rightarrow\infty}h\left(x\right) lim x → ∞ h ( x ) = − 2 =-2 = − 2 lim ⁡ x → − ∞ h ( x ) \lim_{x\rightarrow-\infty}h\left(x\right) lim x → − ∞ h ( x ) = − 2 =-2 = − 2 Horizontal asymptotes: y = − 1 2 y=-\frac12 y = − 2 1 and y = 1 2 y=\frac12 y = 2 1

lim ⁡ x → ∞ h ( x ) \lim_{x\rightarrow\infty}h\left(x\right) lim x → ∞ h ( x ) = 1 =1 = 1 lim ⁡ x → − ∞ h ( x ) \lim_{x\rightarrow-\infty}h\left(x\right) lim x → − ∞ h ( x ) = 0 =0 = 0 No horizontal asymptote

Calculate the following limits and identify the horizontal asympt...

A

$\lim_{x\rightarrow\infty}h\left(x\right)$ $=0$

$\lim_{x\rightarrow-\infty}h\left(x\right)$ $=0$

Horizontal asymptote: $y=0$

B

$\lim_{x\rightarrow\infty}h\left(x\right)$ $=1$

$\lim_{x\rightarrow-\infty}h\left(x\right)$ $=-1$

No horizontal asymptote

C

$\lim_{x\rightarrow\infty}h\left(x\right)$ $=-2$

$\lim_{x\rightarrow-\infty}h\left(x\right)$ $=-2$

Horizontal asymptotes: $y=-\frac12$ and $y=\frac12$

D

$\lim_{x\rightarrow\infty}h\left(x\right)$ $=1$

$\lim_{x\rightarrow-\infty}h\left(x\right)$ $=0$

No horizontal asymptote

Introduction to Limits: Study with Video Lessons, Practice Problems & Examples