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Expert: - 7/23/2010


Question
I am trying to prove that the limit of a function does not exist with the delta epsilon notation.  The questions are as follows:
lim(x->0)[1/x^2] and lim(x->0)[x + sgn(x)].  

Thanks

Answer
If that is (lim x->0)(1/x˛), then suppose the limit is A.
This means that 1/x˛ - A < delta for x < epsiolon.
If x < epsilon, the 1/x˛ is greater than 1/epsilon˛, and this has no limits any epsilon,
so it can't be shown.

As far as lim(x->0)(x + sgn(x)), what is sgn(x)?
If it is sin(x), then for x smaller then epsilon, take delta to be 2 epsilon.
Thus, y is smaller than delta since x is epsilon and and sin(x) is < epsilon for x = epsilon.

If sgn(x) means the sign of x, as in ±1, there is no limit,
for from the positive side it would be 1 and from the negative side it would be -1.

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