Question

Solve 73 make sure to also define the limits in the parts a and b

Answer 1

73.

`f(x)=(3x^4+3x^3-36x^2)/(x^4-25x^2+144)`

a)

`\lim_(x\to\infty)f(x)=\lim_(x\to\infty)((3+(3)/(x)-(36)/(x^2))/(1-(25)/(x^2)+(144)/(x^4)))=3`
`\lim_(x\to-\infty)f(x)=\lim_(x\to-\infty)((3+(3)/(x)-(36)/(x^2))/(1-(25)/(x^2)+(144)/(x^4)))=3\cdot(1)/(2)=3`

b)

Since we can't divide by zero, we need to find when:

`x^4-2x^2+144=0`

But before, we can factor the numerator and the denominator:

`\begin{gathered} (3x^2(x^2+x-12))/(x^4-25x^2+144)=(3x^2((x+4)(x-3)))/((x-3)(x-3)(x+4)(x+4)) \\ so: \\ (3x^2)/((x+3)(x-4)) \end{gathered}`

Now, we can conclude that the vertical asymptotes are located at:

`\begin{gathered} (x+3)(x-4)=0 \\ so: \\ x=-3 \\ x=4 \end{gathered}`

so, for x = -3:

`\lim_(x\to-3^-)f(x)=\lim_(x\to-3^-)-(162)/(x^4-25x^2+144)=-162(-\infty)=\infty`
`\lim_(x\to-3^+)f(x)=\lim_(x\to-3^+)-(162)/(x^4-25x^2+144)=-162(\infty)=-\infty`

For x = 4:

`\lim_(x\to4^-)f(x)=\lim_(n\to4^-)(384)/(x^4-25x^2+144)=384(-\infty)=-\infty`
`\lim_(x\to4^-)f(x)=\lim_(n\to4^-)(384)/(x^4-25x^2+144)=384(-\infty)=-\infty`