anonymous
  • anonymous
limit as x approaches zero of quantity negative six plus x divided by x to the fourth power.
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
anonymous
  • anonymous
\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~0}\frac{6+x}{x^4}}\) this?
anonymous
  • anonymous
yes!

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anonymous
  • anonymous
actually negative 6
anonymous
  • 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
anonymous
  • anonymous
wait, approaches -6?
anonymous
  • anonymous
oh, -6+x! Doesn't matter
anonymous
  • anonymous
you would be then getting \(\large \displaystyle \lim_{x \rightarrow ~0}\frac{-6}{x^4}+\frac{x}{x^4}\) and still DNE
anonymous
  • anonymous
oh okay thanks!
anonymous
  • anonymous
Wait so which one is it?
anonymous
  • anonymous
it is infinity not zero, i think
Vocaloid
  • Vocaloid
I don't believe L'Hopital's rule applies, -6/0 doesn't count as an indeterminate form
anonymous
  • anonymous
no she meant part 2, but there x's xcancel
anonymous
  • anonymous
What is that rule though?
anonymous
  • 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
anonymous
  • anonymous
this is L'Hospital's Rule
anonymous
  • anonymous
Could you guys help me with this?
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anonymous
  • 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?)
anonymous
  • 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)
Plasmataco
  • Plasmataco
Well x=2 is an asymptote
Vocaloid
  • Vocaloid
@phunish well, it's asking for the limit as the function approaches x = 2 from the left, any ideas?
anonymous
  • anonymous
ok, you see that the graph has two parts (two sticks :D) right/
anonymous
  • anonymous
?

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