Question

Prove:
lim x^3 = 8.
x approaches 2

Answer 1

simply plug 2 into x^3, and you will see that as x approaches 2 the limit is 8

Answer 2

can u write a proof is basically what im asking using epsilon and delta

Answer 3

ugh, i can........

Answer 4

THanks...can u..tahts how my teacher wants it

Answer 5

\[if: abs(x-2) < \delta\]
\[then: abs(x^3 - 8) < \epsilon\]
\[-\epsilon < (x^3 - 8) < \epsilon\]
\[-\epsilon + 8 < (x^3) < \epsilon + 8\]
\[-cuberoot(\epsilon + 8) < x < cuberoot(\epsilon + 8)\]
\[(-cuberoot(\epsilon + 8) - 2) < x - 2 < (cuberoot(\epsilon + 8) - 2)\]
\[abs(x - 2) < (cuberoot(\epsilon + 8) - 2) \]
\[ set \delta = (cuberoot(\epsilon + 8) - 2)\]

Answer 6

how did u introduce epsilon into the problem like from teh if to teh then statement? or why is taht true?

Answer 7

The statement:


has the following precise definition. Given any real number\[\epsilon>0\] there exists another real number \[\delta>0\] so that:
if:\[0<\left| x-a \right|<\delta, then \left| f(x)-L \right|<\epsilon\]