## anonymous one year ago limit as x approaches zero of quantity negative six plus x divided by x to the fourth power.

1. anonymous

@Vocaloid

2. anonymous

$$\large\color{slate}{\displaystyle\lim_{x \rightarrow ~0}\frac{6+x}{x^4}}$$ this?

3. anonymous

yes!

4. anonymous

actually negative 6

5. anonymous

$$\large\color{slate}{\displaystyle\lim_{x \rightarrow ~0}\frac{6+x}{x^4}}$$ $$\large\color{slate}{\displaystyle\lim_{x \rightarrow ~0}\frac{6}{x^4}+\lim_{x \rightarrow ~0}\frac{x}{x^4}}$$ so I don't think you will get anything defined out of the limit

6. anonymous

wait, approaches -6?

7. anonymous

oh, -6+x! Doesn't matter

8. anonymous

you would be then getting $$\large \displaystyle \lim_{x \rightarrow ~0}\frac{-6}{x^4}+\frac{x}{x^4}$$ and still DNE

9. anonymous

oh okay thanks!

10. anonymous

Wait so which one is it?

11. anonymous

it is infinity not zero, i think

12. Vocaloid

I don't believe L'Hopital's rule applies, -6/0 doesn't count as an indeterminate form

13. anonymous

no she meant part 2, but there x's xcancel

14. anonymous

What is that rule though?

15. anonymous

that is to differentiate top and bottom, IF you get 0/0 or ∞/∞, when you plug in the value that x approaches into the limit

16. anonymous

this is L'Hospital's Rule

17. anonymous

Could you guys help me with this?

18. anonymous

So for example, I have: $$\large \displaystyle \lim_{x \rightarrow ~0}\frac{\sin(x)}{x}$$ and there you would apply this rule (can you tell me why?)

19. anonymous

i don't see a reason for one-sided limit not to exist (unless the function is totally not on that interval, or if it goes into infinity - asymptote)

20. Plasmataco

Well x=2 is an asymptote

21. Vocaloid

@phunish well, it's asking for the limit as the function approaches x = 2 from the left, any ideas?

22. anonymous

ok, you see that the graph has two parts (two sticks :D) right/

23. anonymous

?