How do you use the epsilon delta definition to prove that the limit of #x^2-4# as #x->2#?

Answer 1

#lim_(x->2)(x^2-4) = 0#

The #epsilon - delta# (epsilon - delta) definition of a limit is as follows:
The limit of #f(x)# as #x# approaches #a# is #L#, denoted #lim_(x->a)f(x)=L# if for every #epsilon > 0# there exists a #delta > 0# such that #0 < |x-a| < delta# implies #|f(x) - L| < epsilon#.
In more intuitive terms, we have #lim_(x->a)f(x) = L# if we can make #f(x)# arbitrarily close to #L# by making #x# close to #a#.
Using that definition, we can show that #lim_(x->2)(x^2-4) = 0#

Proof:

Let #epsilon > 0# be arbitrary, and let #delta = sqrt(4+epsilon)-2#
Note that #delta > 0#, and that if #0 < |x-2| < delta#, we have
#-delta < x-2 < delta# #=> -delta + 4 < x+2 < delta + 4# #=> -delta - 4 < x+2 < delta + 4# #=> |x+2| < delta + 4#
Now, suppose #0 < |x-2| < delta#. Then
#|(x^2-4)-0| = |x^2-4|#
#=|x-2|*|x+2|#
# < delta(delta+4)#
#=delta^2+4delta#
#=(sqrt(4+epsilon)-2)^2+4(sqrt(4+epsilon)-2)#
#=4+epsilon-4sqrt(4+epsilon)+4+4sqrt(4+epsilon)-8#
#=epsilon#
We have shown that #0 < |x-2| < delta# implies #|(x^2-4)-0| < epsilon#, and thus #lim_(x->2)(x^2-4) = 0# by the #epsilon-delta# definition of a limit. ∎
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Answer 2
To use the epsilon-delta definition to prove the limit of a function, we need to show that for any given epsilon (ε) greater than zero, there exists a delta (δ) greater than zero such that if the distance between x and 2 (|x - 2|) is less than delta, then the distance between f(x) and the limit L (|f(x) - L|) is less than epsilon. In this case, we want to prove that the limit of f(x) = x^2 - 4 as x approaches 2. Let's proceed with the proof: Given ε > 0, we need to find a δ > 0 such that if 0 < |x - 2| < δ, then |(x^2 - 4) - L| < ε. Let's start by manipulating the expression |(x^2 - 4) - L|: |(x^2 - 4) - L| = |x^2 - 4 - L| = |x^2 - (L + 4)| Now, we can try to bound this expression by using the difference of squares: |x^2 - (L + 4)| = |(x - 2)(x + 2) - (L + 4)| = |(x - 2)(x + 2) - (L + 4)(x - 2)/(x - 2)| Expanding the numerator and simplifying, we get: |(x - 2)(x + 2) - (L + 4)(x - 2)/(x - 2)| = |(x - 2)(x + 2 - (L + 4)/(x - 2))| Now, we can see that if we choose δ such that |x - 2| < δ, then we can also ensure that |x - 2| ≠ 0, which allows us to divide by (x - 2) without any issues. Let's choose δ = min(1, ε/5). Now, if |x - 2| < δ, we have: |x - 2| < δ ≤ 1, and ε/5 > 0 |x - 2| < 1 implies -1 < x - 2 < 1, which further implies 1 < x + 2 < 3 Hence, we can bound the expression as follows: |(x - 2)(x + 2 - (L + 4)/(x - 2))| ≤ |(x - 2)(3 - (L + 4)/(x - 2))| Now, let's choose a value for δ such that |(x - 2)(3 - (L + 4)/(x - 2))| < ε: |(x - 2)(3 - (L + 4)/(x - 2))| < ε Since |x - 2| < δ ≤ 1, we can assume that |x - 2| ≠ 0, allowing us to divide by (x - 2) without any issues. Hence, we have: |(x - 2)(3 - (L + 4)/(x - 2))| = |3(x - 2) - (L + 4)| = |3x - 6 - L - 4| = |3x - L - 10| Now, we want to choose δ such that |3x - L - 10| < ε. Let's choose δ = ε/3. If |x - 2| < δ = ε/3, then we have: |3x - L - 10| = 3|x - 2| < 3(ε/3) = ε Therefore, we have shown that for any ε > 0, we can choose δ = min(1, ε/5) such that if 0 < |x - 2| < δ, then |(x^2 - 4) - L| < ε. This satisfies the epsilon-delta definition of the limit, proving that the limit of f(x) = x^2 - 4 as x approaches 2 is L.
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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