Let's take any positive real number epsilon.
Define the positive delta depending on the epsilon as following.
delta = min((1/2)epsilon|a|^2, (1/2)|a|)
For any x satisfying |x-a| < delta, |1/x - 1/a|=|(a-x) / (ax)|
=|a-x| / |a||x|.
Since lx-a| < delta <= 2epsilon|a|^2, |a-x| < (1/2)epsilon|a|^2.
Since |x-a| < delta <= (1/2)|a|, a-(1/2)|a| < x < a+(1/2)|a|.
If a>0, then (1/2)|a| < x < (3/2)|a|.
If a<0, then (-3/2)|a| < x < (-1/2)|a|.
Hence, (1/2)|a| < |x| < (3/2)|a| providing that a is not zero.
It implies that 1/|x| < 2/|a|.
Therefore, |a-x| / |a||x| < epsilon|a|^2 / |a|^2 = epsilon.
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