LaddiusMaximus
  • LaddiusMaximus
lim x->0 ((1=x)^3-1/x)
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
1=x means 1+x?
LaddiusMaximus
  • LaddiusMaximus
crap! yeah
anonymous
  • anonymous
\[(1+x)^3-\frac{1}{x}\]so it looks like

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LaddiusMaximus
  • LaddiusMaximus
what?
anonymous
  • anonymous
is that how the equation is written?
LaddiusMaximus
  • LaddiusMaximus
no it is written (1+x)^3-1 / x\[\left( 1+x \right)^{3} -1\div x\]
anonymous
  • anonymous
\[\frac{(1+x)^3-1}{x}\]so
LaddiusMaximus
  • LaddiusMaximus
correct
anonymous
  • anonymous
ok
anonymous
  • anonymous
let me type it out
anonymous
  • anonymous
\[\lim_{x \rightarrow 0} \frac{(x+1-1)[(x+1)^2+(x+1)+1]}{x}\]
anonymous
  • anonymous
That's because of a difference of 2 cubes
anonymous
  • anonymous
\[\lim_{x \rightarrow 0} \frac{x(x^2+2x+1+x+2)}{x}=\lim_{x \rightarrow 0} x^2+3x+3\]
anonymous
  • anonymous
plug in 0 for x from here, there's your solution
anonymous
  • anonymous
Understand?
anonymous
  • anonymous
The limit is 3
LaddiusMaximus
  • LaddiusMaximus
got you. Thank you!

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