Epsilon-Delta Proofs for Limit

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Recall

Very loosing speaking, a limit is the value to which a mathematical function gets closer and closer to as the input gets closer and closer to some given value.

A limit describe what is happening around a given point, say “a”. It is the value that the function approaches as the input approaches “a”, and it does not depend on the actual value of the function at a, or even on whether the function is defined at “a” at all.

Limits are essential to calculus and mathematical analysis and the understanding of how functions behave. The concept of a limit can be written or expressed as $\lim_{x \to a} f(x) = L.$ This notation is read as “the limit of f as x approaches a equals L”.

Intuitively, this means that the values of f(x) can be made arbitrarily close to L (and I mean as close as we like, e.g., L ± 0.1, L ± 0.01, L ± 0.001, and so on), by choosing values of x sufficiently close to a, but not necessarily equal to a.

Formal definition. We say that the limit of f, as x approaches a, is L, and write $\lim_{x \to a}f(x) = L$. For every real ε > 0, there exists a real δ > 0 such that whenever 0 < | x − a | < δ we have | f(x) − L | < ε. In other words, we can make f(x) arbitrarily close to L, f(x)∈ (L-ε, L+ε) (within any distance ε > 0) by making x sufficiently close to a (within some distance δ > 0, but not equal to a) (x ∈ (a-δ, a+δ), x ≠ a) -Fig 1.a.-

Graphically, one can draw a horizontal band (L − ε, L + ε) around L; there should exist a vertical interval or band (a − δ, a + δ) (punctured at or excluding a) such that the graph of f in that vertical strip (except possibly a) lies entirely inside the given horizontal band.

Solved examples with epsilon-delta proofs

• Prove that $\lim_{x \to 0} x^{2} = 0$ -1.b.-

$\forall \epsilon>0, \exists \delta>0: 0<|x|<\delta\ \implies |x^{2}|<\epsilon$. We need $|x^{2}|<\epsilon$. This is equivalent to $|x| \lt \sqrt{\epsilon}.$

Let’s choose $\delta = \sqrt \epsilon$, then $\forall \epsilon>0, \exists \delta>0: 0<|x|<\sqrt \epsilon, \implies |x^{2}|<\epsilon$

That’s true because $|x^{2}-0|=|x|^{2}<(\sqrt \epsilon)^{2} = \epsilon$

• Prove that $\lim_{x \to 2} 5x - 4 = 6$ -1.c.-

$\forall \epsilon>0, \exists \delta>0: 0<|x-2|<\delta \implies |5x - 4 -6|<\epsilon$. Simplify the expression: |(5x −4) − 6∣ = ∣5x − 10∣ = 5∣x − 2∣. We require $5∣x − 2∣ <\epsilon$, i.e, $∣x − 2∣ <\frac{\epsilon}{5}$

Let’s choose $\delta = \frac {\epsilon}{5}$, then $\forall \epsilon>0, \exists \delta>0: 0<|x-2|<\frac {\epsilon}{5} \implies ∣(5x −4) −6∣ = 5|x - 2|<\frac {5\epsilon}{5} = \epsilon$

• Prove that $\lim_{x \to 4} (2x -3) = 5$

$\forall \epsilon>0, \exists \delta>0: 0<|x-4|<\delta \implies |2x - 3 -5|<\epsilon$

Simplify: |2x - 3 -5| = |2x -8| = 2|x -4|. We need $2|x-4| \lt \epsilon$. Choose δ = ε/2.

Let ε >0 be given. Choose δ = ε/2. Suppose 0 < ∣x − 4∣ < δ, then $|2x - 3 -5| = 2|x-4| < 2 \cdot \frac{\epsilon}{2} = \epsilon$

• $\lim_{x \to 5} (\frac{1}{x-3}) = \frac{1}{2}$

$\forall \epsilon>0, \exists \delta>0: 0<|x-5|<\delta,\implies |\frac{1}{x-3} -\frac{1}{2}|<\epsilon$

$|\frac{1}{x-3} -\frac{1}{2}| = |\frac{2-x+3}{2(x-3)}| = |\frac{-x+5}{2(x-3)}| = \frac{1}{2}·|x-5|·\frac{1}{|x-3|}$ 🚀

We know that $0<|x-5|<\delta$. Let’s choose δ < 1, |x-5| < δ < 1 ↭ 4 < x < 6, hence 1 < x -3 < 3, so |x -3| =[x - 3 > 1 > 0] x - 3 > 1, and $\frac{1}{|x-3|} \lt 1$.

[🚀] $|\frac{1}{x-3} -\frac{1}{2}| = \frac{1}{2}·|x-5|·\frac{1}{|x-3|} < \frac{1}{2}·δ·1$ [So we pick δ = min{1, 2ε}] ≤ $\frac{1}{2}·2ε·1 = ε$∎

• $\lim_{x \to 0} xsin(\frac {1}{x}) = 0.$

$\forall \epsilon>0, \exists \delta>0: 0<|x|<\delta,\implies |xsin(\frac {1}{x})|<\epsilon$

Let’s choose δ = ε, then $0<|x|<\delta \implies |xsin(\frac {1}{x})| = |x|·|sin(\frac {1}{x})| ≤[\text{Use the bound |sin(θ)| ≤ 1 for all real θ}] |x|< δ = ε$

• $\lim_{x \to 2} 3x^2 = 12$

$\forall \epsilon>0, \exists \delta>0: 0<|x-2|<\delta \implies |3x^2 -12|<\epsilon$

$|3x^2 -12| = 3|x^2-4| = 3|(x-2)(x+2)| = 3|x-2||x+2|$ < [🚀]

If |x-2| < 1, then -1 < x -2 < 1 ⇒[Let’s add 4] 3 < x + 2 < 5 ⇒ -5 < 3 < x + 2 < 5 ⇒ |x + 2| < 5

< [🚀] 3·δ·5 <[Let’s choose δ = min{1, ε/15}] ε∎

• $\lim_{x \to 5} \sqrt{3x+1} = 4$

$\forall \epsilon>0, \exists \delta>0: 0<|x-5|<\delta,\implies |\sqrt{3x+1} - 4|<\epsilon$

$|\sqrt{3x+1} - 4| = |\frac{(\sqrt{3x+1} - 4)(\sqrt{3x+1} + 4)}{\sqrt{3x+1} + 4}| = |\frac{3x+1-16}{\sqrt{3x+1} + 4}| = |\frac{3x-15}{\sqrt{3x+1} + 4}|$=[Choose δ < 1, x is between 4 and 6 and the denominator is not only positive, but greater than 1] $\frac{3|x-5|}{\sqrt{3x+1} + 4}$ [$\sqrt{3x+1} + 4>1 ⇒ \frac{1}{\sqrt{3x+1} + 4} < 1$] < 3|x-5| < 3δ ≤[Choose δ = min{1, ε/3}] ε ∎

• $\lim_{x \to -2} -3x+5 = 11$

$\forall \epsilon>0, \exists \delta>0: 0<|x+2|<\delta,\implies |-3x+5 - 11|<\epsilon$

|-3x+5 - 11| = |-3x -6| = |-3(x + 2)| = 3|x+2| < 3·δ =[Choose δ = ^ε⁄₃] ε∎

• $\lim_{x \to 3} \frac{x}{x-2} = 3$

$\forall \epsilon>0, \exists \delta>0: 0<|x-3|<\delta,\implies |\frac{x}{x-2} - 3|<\epsilon$

$|\frac{x}{x-2} - 3| = |\frac{x-3x+6}{x-2}| = |\frac{-2x+6}{x-2}|=2|x-3|\frac{1}{|x-2|} < 2δ\frac{1}{|x-2|}$ [🚀]

|x-3|<δ ⇒[Choose δ = min{1/2, something else}] $\frac{-1}{2}< x-3 < \frac{1}{2}$ ⇒[Add 1] $\frac{1}{2}< x-2 < \frac{3}{2} ⇒ 2 > \frac{1}{x-2} > \frac{2}{3} > -2 ⇒ \frac{1}{|x-2|}< 2$

When you take reciprocals, the inequalities reverse if the quantities are positive. For positive numbers, $a \lt b \implies \frac{1}{a} \gt \frac{1}{b}$

[🚀] $|\frac{x}{x-2} - 3|$ < 2·δ·2 ≤[Choose δ = min{1/2, ε/4}] ε∎

• $\lim_{x \to 2} x^3 = 8$

$\forall \epsilon>0, \exists \delta>0: 0<|x-2|<\delta,\implies |x^3 - 8|<\epsilon$

$|x^3 - 8| = |x-2|·|x^2+2x+4|< δ·|x^2+2x+4|$ [🚀]

Choose δ = min{1, something else}, |x-2|< 1 ↭ -1 < x -2 < 1 ⇒[Adding 2] 1 < x < 3 ⇒ -3 < 1 < x < 3 ⇒ |x| < 3.

$|x^3 - 8| 🚀< δ·|x^2+2x+4|≤$ [By the triangle inequality, |a+b| ≤ |a| + |b|] δ·(|x|²+ 2|x| + |4|) < δ(3² +2·3 + 4) = δ·19 ≤[Choose δ = min{1, ε/19}] ε ∎