Answer to Question #236202 in Math for Fia

Question

Determine the Vertical Asymptote

f(x) = 1-x / x^3 -64

f(x) = 7x + 2/ x^2+1

Answer 1

1.

$$f(x)=\dfrac{1-x}{x^3-64}, x\in \R$$

$$x^3-64\not=0$$

$$(x-4)(x^2+x+16)\not=0$$

$$x\not=4$$

$$\lim\limits_{x\to4^-}f(x)=\lim\limits_{x\to4^-}\dfrac{1-x}{x^3-64}$$

$$=\lim\limits_{x\to4^-}\dfrac{1-x}{(x-4)(x^2+x+16)}=\infin$$

$$\lim\limits_{x\to4^+}f(x)=\lim\limits_{x\to4^+}\dfrac{1-x}{x^3-64}$$

$$=\lim\limits_{x\to4^+}\dfrac{1-x}{(x-4)(x^2+x+16)}=-\infin$$

Vertical asympote: $x=4.$

2.

$$f(x)=\dfrac{7x+2}{x^2+1}, x\in \R$$

$$x^2+1>0, x\in\R$$

There is no vertical asymptote.