Find, and then prove, each limit, using the epsilon-delta (ε-ẟ) definition:
These problems are from the video on blackpenredpen's YouTube channel
linked here: youtube.com/watch?v=luiJmucU7lI and these answers are based on
blackpenredpen's technique from his YouTube channel, a video of an example
[ of which is linked here: youtube.com/watch?v=DdtEQk_DHQs.
1. lim as x -> 1 of 6x - 2 = 6*(1) - 2 = 6 - 2 = 4, because
if we let f(x) = 6x - 2, L = 4, and a = 1, then we have the following:
Given ε > 0, choose ẟ = ε/6. Suppose 0 < |x - a| = |x - 1| < ẟ,
and check: |f(x) - L| = |(6x - 2) - 4| = |6x - 6| = |6 * (x - 1)|,
so we now have |f(x) - L| = |6| * |x - 1| < 6 * ẟ = 6 * (ε/6) = ε. [[QED]
2. lim as x -> -2 of (2x + 7) = 2*(-2) + 7 = -4 + 7 = 3, because
if we let f(x) = 2x + 7, L = 3, and a = -2, then we have the following:
Given ε > 0, choose ẟ = ε/2. Suppose 0 < |x - a| = |x - (-2)| = |x + 2| < ẟ,
and check: |f(x) - L| = |(2x + 7) - 3| = |2x + 4| = |2 * (x + 2)|,
so we now have |f(x) - L| = |2| * |x + 2| < 2 * ẟ = 2 * (ε/2) = ε. [QED]
3. lim as x -> 3 of (2x/3 - 5) = 2*(3)/3 - 5 = 2 - 5 = -3, because
if we let f(x) = 2x/3 - 5, L = -3, and a = 3, then we have the following:
Given ε > 0, choose ẟ = 3ε/2. Suppose 0 < |x - a| = |x - 3| < ẟ,
and check: |f(x) - L| = |(2x/3 - 5) - (-3)| = |2x/3 - 2| = |2/3 * (x - 3)|,
so we now have |f(x) - L| = |2/3| * |x - 3| < 2/3 * ẟ = 2/3 * (3ε/2) = ε. [QED]
4. lim as x -> 4 of √(x) = √(4) = 2, because
if we let f(x) = √(x), L = 2, and a = 4, then we have the following:
Given ε > 0, choose ẟ = ε. Suppose 0 < |x - a| = |x - 4| < ẟ, and
check: |f(x) - L| = |√(x) - 2| = |√(x) - 2| * |√(x) + 2| / |√(x) + 2|,
so we now have |f(x) - L| = |x - 4| / |√(x) + 2| < |x - 4| < ẟ = ε,
because |√(x) + 2| > 1 for all x >= 0. [QED]
5. lim as x -> 6 of √(3x - 2) = √(3*(6) - 2) = √(18 - 2) = √(16) = 4, because
if we let f(x) = √(3x - 2), L = 4, and a = 6, then we have the following:
Given ε > 0, choose ẟ = ε/3. Suppose 0 < |x - a| = |x - 6| < ẟ, and
check: |f(x) - L| = |√(3x - 2) - 4| = |√(3x - 2) - 4| * |√(3x - 2) + 4| / A,
where A = |√(3x - 2) + 4|, which is > 1 for all 3x - 2 > 0, so we now have
|f(x) - L| = |3x - 2 - 16| / A < |3x - 18| = 3|x - 6| < 3ẟ = 3*(ε/3) = ε. [QED]
6. lim as x -> 3 of (2x² + 1) = 2*(3)² + 1 = 2*9 + 1 = 18 + 1 = 19, because
if we let f(x) = 2x² + 1, L = 19, and a = 3, then we have the following:
Given ε > 0, choose ẟ = min(1, ε/14). Suppose 0 < |x - a| = |x - 3| < ẟ,
and check: |f(x) - L| = |(2x² + 1) - 19| = |2x² - 18| = 2 * |x + 3| * |x - 3|,
so if we let ẟ be <= 1, then |x - 3| < ẟ means that |x - 3| < 1, so we have:
-1 < x - 3 < 1, so 5 < x + 3 < 7, so |x + 3| < 7, so we now have:
|f(x) - L| = 2 * |x + 3| * |x - 3| < 2*7*ẟ = 14 * ẟ <= 14 * (ε/14) = ε. [QED]
7. lim as x -> -2 of (x² - 3x) = (-2)² - 3*(-2) = 4 - (-6) = 4 + 6 = 10, because
if we let f(x) = x² - 3x, L = 10, and a = -2, then we have the following:
Given ε > 0, choose ẟ = min(1, ε/8).
Suppose 0 < |x - a| = |x - (-2)| = |x + 2| < ẟ, and check:
|f(x) - L| = |(x² - 3x) - 10| = |x - 5| * |x + 2|, so if we let ẟ be <= 1,
then |x + 2| < ẟ means that |x + 2| < 1, so we have:
-1 < x + 2 < 1, so -8 < x - 5 < -6, so |x - 5| < 8, so we now have:
|f(x) - L| = |x - 5| * |x + 2| < 8 * ẟ <= 8 * (ε/8) = ε. [QED]
8. lim as x -> 2 of (x³) = (2)³ = 8, because
if we let f(x) = x³, L = 8, and a = 2, then we have the following:
Given ε > 0, choose ẟ = min(1, ε/19). Suppose 0 < |x - a| = |x - 2| < ẟ,
and check: |f(x) - L| = |x³ - 8| = |x² + 2x + 4| * |x - 2| <=
(|x²| + |2x + 4|) * |x - 2| = (|x²| + 2 * |x + 2|) * |x - 2|,
so if we let ẟ be <= 1, then |x - 2| < ẟ means that |x - 2| < 1, so we have:
-1 < x - 2 < 1, so 1 < x < 3 and 3 < x + 2 < 5, so |x²| < 3² = 9 and
|x + 2| < 5, so we now have: |f(x) - L| = (|x²| + 2 * |x + 2|) * |x - 2| <
(9 + 2 * 5) * ẟ = (9 + 10) * ẟ = 19 * ẟ <= 19 * (ε/19) = ε. [QED]
9. lim as x -> 2 of (1 / x) = 1/2, because:
if we let f(x) = 1/x, L = 1/2, and a = 2, then we have the following:
Given ε > 0, choose ẟ = min(1, ε). Suppose 0 < |x - a| = |x - 2| < ẟ, & check:
|f(x) - L| = |1/x - 1/2| = |2/(2x) - x/(2x)| = |2 - x| / |2x| = |x - 2| / |2x|,
so if we let ẟ be <= 1, then |x - 2| < ẟ means that |x - 2| < 1, so we have:
-1 < x - 2 < 1, so 1 < x < 3, so 2 < 2x < 6, so |2x| > 1, so we now have:
|f(x) - L| = |x - 2| / |2x| < |x - 2| < ẟ = ε. [QED]
10. lim as x -> 1 of 1 / (2x + 1) = 1 / (2*(1) + 1) = 1 / (2 + 1) = 1/3, because
if we let f(x) = 1 / (2x + 1), L = 1/3, and a = 1, then we have the following:
Given ε > 0, choose ẟ = min(1, 3ε/2). Suppose 0 < |x - a| = |x - 1| < ẟ, and
check: |f(x) - L| = |1/(2x+1) - 1/3| = |3/[3*(2x+1)] - (2x+1)/[3*(2x+1)]|
= |3 - (2x+1)| / [3*|2x+1|] = |2 - 2x| / [3 * |2x + 1|]
= 2 * |1 - x| / [3 * (2x + 1)] = 2 * |x - 1| / [3 * |2x + 1|],
so if we let ẟ be <= 1, then |x - 1| < ẟ means that |x - 1| < 1, so we have:
-1 < x - 1 < 1, so 0 < x < 2, so 0 < 2x < 4, so 1 < 2x + 1 < 5,
so |2x + 1| > 1, so we now have:
|f(x) - L| = 2 * |x - 1| / [3 * |2x + 1|] < 2 * ẟ/3 <= 2 * (3ε/2)/3 = ε. [QED]
