How do you prove that the limit of #f(x) = 2 - (1/x)# as x approaches 1 using the epsilon delta proof?

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Answer 1 · 2016-12-02T11:24:14

By choosing in the proof: #delta = epsilon/(1+epsilon)#

We start noting that #f(x)# is continuous in #x=1#, so that:

#lim_(x->1) 2-1/x = 2-1/1=1#

The formal #epsilon - delta # proof of this limit is as follows:

#AA epsilon > 0, EE delta > 0# such that :

#|2-1/x-1| < epsilon# for #x in (1-delta, 1+delta)#

Note that:

#|2-1/x-1| = |1-1/x| = |(x-1)/x|= |x-1|/|x|#

So, given #epsilon#, choose #delta = epsilon/(1+epsilon)#

For #x in (1-delta, 1+delta)# evidently:

#|x-1| < delta#

and

#|x| > 1 -delta#,

so:

# |x-1|/|x| < delta/(1-delta) = (epsilon/(1+epsilon))/(1-(epsilon/(1+epsilon)))= epsilon#

Q.E.D.