Verify the following limit?

Question

Use the $epsilon − delta$ definition of a limit to verify the following limit:

$lim_(x→3) (x)/(6-x)= 1$

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Answer 1 · 2018-02-27T21:21:05

Evaluate the difference:

$abs(x/(6-x)-1) = abs ( ( x-6+x)/(6-x)) = abs ((2x-6)/(6-x)) = 2 abs (x-3)/abs(x-6)$

Given now $epsilon > 0$ choose $delta_epsilon < min(1,epsilon)$

For $x in (3-delta_epsilon,3+delta_epsilon)$ then we have:

$abs (x-3) < delta_epsilon$

$3-delta_epsilon < abs(x-6) < 3+delta_epsilon$

then as $delta_epsilon < 1$

$abs(x-6) > 2$

So for $x in (3-delta_epsilon,3+delta_epsilon)$ :

$abs(x/(6-x)-1) < (2delta_epsilon)/ 2$

$abs(x/(6-x)-1) < delta_epsilon$

and as $delta_epsilon < epsilon$ :

$x in (3-delta_epsilon,3+delta_epsilon) => abs(x/(6-x)-1) < epsilon$

which proves the limit.