Sorry about the vague title. I've made apost similar to this... Here it goes:

Give an epsilon - N proof that:

if a[n] -> a then a[n]^2 -> a^2

and the same for

if a[n] -> a then a[n]^4 -> a^4

Any Ideas? Again, cheers

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- May 20th 2009, 03:16 PMmitch_nufcRelations between similar limits?
Sorry about the vague title. I've made apost similar to this... Here it goes:

Give an epsilon - N proof that:

if a[n] -> a then a[n]^2 -> a^2

and the same for

if a[n] -> a then a[n]^4 -> a^4

Any Ideas? Again, cheers - May 20th 2009, 10:10 PMGamma
Let $\displaystyle \epsilon > 0$ be given.

By assumption $\displaystyle a_n \rightarrow a < \infty$.

Choose N such that for all n>N

$\displaystyle |a_n-a|<\frac{\epsilon}{2a}$

Now for the real N, we may need to go bigger depending on the epsilon, we must consider |M| is the maximum value of the sequence past N, at some point the sequence must be bounded if it converges to a. The point is we need to choose N so that $\displaystyle |a_n-a|<min\{\frac{\epsilon}{2a}, \frac{\epsilon}{2|M|}\}$ for all n>N.

$\displaystyle |a_n^2 - a^2|= |a_n^2 - a_na + a_na -a^2| \leq ||a_n||a_n-a| + |a||a-a_n|< |M|\frac{\epsilon}{2|M|} + a\frac{\epsilon}{2a}=\epsilon$

I think you can figure out the second part, remember

$\displaystyle a^4=(a^2)^2$