Find the limit of the function algebraically.

limit as x approaches zero of quantity x cubed plus one divided by x to the fifth power.

Question

Find the limit of the function algebraically.

limit as x approaches zero of quantity x cubed plus one divided by x to the fifth power.

anonymous · Answer

please help!!

amoodarya · Answer

$$\lim_{x \rightarrow 0}\frac{ x^3+1 }{ x^5 }$$
did you mean it ?

anonymous · Answer

yes

amoodarya · Answer

$$\lim_{x \rightarrow 0^+}\frac{ x^3+1 }{ x^5 }=\frac{ 1 }{ 0^+ }=+ \infty\lim_{x \rightarrow 0^-}\frac{ x^3+1 }{ x^5 }=\frac{ 0^-+1 }{ 0^- }=+ \infty$$

amoodarya · Answer

sorry second was -inf

anonymous · Answer

okay so let me show you the answer choices

amoodarya · Answer

$$\lim_{x \rightarrow 0^-}\frac{ x^3+1 }{ x^5 }=\frac{ 0^-1+1 }{ 0^- }=\frac{ 1 }{ 0^- }=- \infty$$

anonymous · Answer

i think i know which one it might be

anonymous · Answer

0

 -9

 Does not exist

 9

amoodarya · Answer

Does not exist

anonymous · Answer

okay i figured that, thanks so much!!

anonymous · Answer

do you have time for one more

amoodarya · Answer

write it

anonymous · Answer

Find the derivative of f(x) = negative 9 divided by x at x = -8

amoodarya · Answer

$$f(x)=\frac{ -9 }{ x }=?$$

anonymous · Answer

yes

anonymous · Answer

answer choices:  8 divided by 9

 9 divided by 8

 64 divided by 9

 9 divided by 64

amoodarya · Answer

$$f(x)=\frac{ -9 }{ x }\f'(x)=\frac{ 0(x)-(1)(-9) }{x^2 }\f'=\frac{ 9 }{ x^2 }\now\put\x=-8$$

anonymous · Answer

-64?