How do you prove that the limit of #x+2y= 1# as (x,y) approaches (1,0) using the epsilon delta proof?

2 Answers
Konstantinos Michailidis
Dec 4, 2016

Set #y=f(x)=1/2*(1-x)# hence we want show that

#lim_(x->1) f(x)=0#

Suppose we are given # ε > 0#. We then wish to find #δ > 0# such
that for any #x ∈ R#, #0 < |x − 1| < δ# implies #|1/2(1-x) − 0| < ε# or
#1/2*|(x-1)| < ε# or #|(x-1)| <2* ε# .

If we choose #δ=min{2*ε,1}# would be an alright choice for # δ#

Jim H
Dec 17, 2016

If I ever did this it was years ago. I did look at some resources for #epsilon#-#delta# definitions of such limits and here is the result.

Explanation:

Given #epsilon > 0#, choose #delta = epsilon/3#. (Note that #delta# is positive.)

Now if #(x,y)# is within #delta# of #(1,0)# but distinct from #(1,0)#,

then #0 < sqrt((x-1)^2+y^2) < delta#.

So #sqrt((x-1)^2) = abs(x-1) < delta# and #sqrt(y^2) = abs(y) < delta#.

Consequently,

#abs((x+2y)-1)) = abs((x-1)+2y)#

# <= abs(x-1)+2abs(y)#

# < delta + 2delta#

# = epsilon/3 + 2epsilon/3#

# = epsilon#.

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