Question

Limit definition: (Medals shall be awarded)

Calculate using the limit definition:
(x^3) - x

Answer 1

Hm.. Perhaps i'm doing a fool out of myself right now. But eh.. this function is continuous.. so the only useful limits I could think of are
\[ \lim_{x \to \infty}{x^3 - x} = \infty^3 - x = \infty\]
and
\[ \lim_{x \to -\infty}{x^3 - x} = (-\infty)^3 - x = -\infty \]
Is that what you meant? maybe more clarifications? =s
Perhaps you meant using derivative definition? I'm confused. sorry

Answer 2

Yes, i did mean the derivative definition.

Answer 3

The equation that has, Lim h-->0

Answer 4

ok, i'll try =S

Answer 5

sorry it's long =|
\[
f(x) = x^3 - x \\
\lim_{\Delta -> 0}{
\frac{ f(x+\Delta) - f(x) }{\Delta }
} = \\
=\lim_{\Delta -> 0}{
\frac{ (x + \Delta)^3 - (x+\Delta) - (x^3 - x) }{\Delta }
} = \\
=\lim_{\Delta -> 0}{
\frac{ (x + \Delta)^3 - \Delta - x^3 }{\Delta }
} = \\
=\lim_{\Delta -> 0}{
\frac{ (x+\Delta)(x^2 + 2x\Delta + \Delta^2) - \Delta - x^3 }{\Delta }
} = \\
= \lim_{\Delta -> 0}{
\frac{ (x^3 + 2x^2\Delta + x\Delta^2)+(x^2\Delta + 2x\Delta^2 + \Delta^3) - \Delta - x^3 }{\Delta }
} = \\
=\lim_{\Delta -> 0}{
\frac{ 3x^2\Delta + 3x\Delta^2 + \Delta^3 - \Delta }{\Delta }
} = \\
=\lim_{\Delta -> 0}{
\frac{ \Delta(3x^2 + 3x\Delta + \Delta^2 - 1) }{\Delta }
} = \\
=\lim_{\Delta -> 0}{ 3x^2 + 3x\Delta + \Delta^2 - 1 } =
3x^2 + 3x \cdot 0 + 0^2 - 1 = 3x^2 - 1
\]
Btw, I wondered. Is this a riddle or what?

Answer 6

I think I'm still confused by the statement of the problem.
To calculate, doesn't necessarily mean find derivative. Maybe if someone said to calculate the derivative, but I'm not even sure someone would word the problem in such a way.
When you wrote this problem here, is that what it say exactly? To calculate using the limit definition?

Answer 7

Perhaps he just wanted to see if somebody could do this, and didn't take this from anywhere. sounds to me like a riddle idk.

Answer 8

actually, pitamar
you just solved my problem

Answer 9

thank you so much, i was getting the wrong answer

Answer 10

You can also use the alternate form:
\[f'(x)=\lim_{z \rightarrow x} \frac{f(z)-f(x)}{z-x}\]
\[f(x)=x^3-x , f(z)=z^3-z\]
Plug in
\[f'(x)=\lim_{z \rightarrow x}\frac{(z^3-z)-(x^3-x)}{z-x}=\lim_{z \rightarrow x}\frac{z^3-x^3-z+x}{z-x}\]
\[=\lim_{z \rightarrow x}\frac{z^3-x^3}{z-x}-\lim_{z \rightarrow x}\frac{z-x}{z-x}\]
\[=\lim_{z \rightarrow x}\frac{(z-x)(z^2+zx+z^2)}{z-x}-1\]
\[=\lim_{z \rightarrow x}(z^2+xz+x^2)-1\]
z goes to x
so we have
\[(x^2+xx+x^2)-1=x^2+x^2+x^2-1=3x^2-1\]

Answer 11

eww...those z's look like x's

Answer 12

Thank you guys so much