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Definition. A function f is a rule, relationship, or correspondence that assigns to each element of one set (x ∈ D), called the domain, exactly one element of a second set, called the range (y ∈ E).

The pair (x, y) is denoted as y = f(x). Typically, the sets D and E will be both the set of real numbers, ℝ. A mathematical function is like a black box that takes certain input values and generates corresponding output values (Figure E).

Very loosing speaking, a limit is the value to which a function grows close as the input get closer and closer to some other given value.

One would say that the limit of f, as x approaches a, is L, $\lim_{x \to a} f(x)=L$. Formally, for every real ε > 0, there exists a real δ > 0 such that for all real x, 0 < | x − a | < δ implies that | f(x) − L | < ε. In other words, *f(x) gets closer and closer to L, f(x)∈ (L-ε, L+ε), as x moves closer and closer -approaching closer but never touching- to a* (x ∈ (a-δ, a+δ), x≠a)) -Fig 1.a.-

Definition. Let f(x) be a function defined on an interval that contains x = a, **except possibly at x = a**, then we say that, $\lim_{x \to a} f(x) = L$ if

$\forall \epsilon>0, \exists \delta>0: 0<|x-a|<\delta, implies~ |f(x)-L|<\epsilon$

Or

$\forall \epsilon>0, \exists \delta>0: |f(x)-L|<\epsilon, whenever~ 0<|x-a|<\delta$

The squeeze or sandwich theorem. If a function f lies between two functions g and h, and the limits of each of them at a particular point are equal to L, then the limit of f at that particular point is also equal to L.

The theorem is particularly useful when direct substitution or algebraic manipulation is challenging.

Intuitive Explanation:
The Squeeze Theorem helps determine the limit of a function by comparing it to two other functions whose limits are known. **If f(x) is “sandwiched” between g(x) and h(x), and both g(x) and h(x) approach the same limit L as x approaches a certain value, say a, then f(x) must also approach that limit at that point.**

Formal definition. Let f, g, and h be functions defined on an interval I that contains x = a, **except possibly at x = a**. Suppose that for every x in I not equal to a, we have g(x) ≤ f(x) ≤ h(x), and also suppose that $\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L.$ Then, $\lim_{x \to a} f(x) = L.$

Proof:
$\forall \epsilon>0, \exists \delta_1>0: 0<|x-a|<\delta_1,~~implies~~ |g(x)-L|<\epsilon~ *_1$

$\forall \epsilon>0, \exists \delta_2>0: 0<|x-a|<\delta_2,~~implies~~ |h(x)-L|<\epsilon~ *_2$

Let’s choose as $\delta = min(\delta_1, \delta_2)$
$\forall \epsilon>0, \exists \delta>0: 0<|x-a|<\delta,~~implies~~$

$L -\epsilon < g(x) < L +\epsilon$ [*_{1}] and $L -\epsilon < h(x) < L +\epsilon$ [*_{2}]

$L -\epsilon < g(x) ≤ f(x) ≤ h(x) < L +\epsilon ⇨ L -\epsilon < f(x) < L +\epsilon ⇨ |f(x)-L|<\epsilon$ ∎

- $\lim_{x \to 0} f(x),
~~where~~f(x)= x·sin(\frac{1}{x}) = 0$ for x ≠ 0, and f(0) = 0 (Figure i).

$-1 ≤ sin(\frac{1}{x}) ≤ 1 ⇨ -x ≤ xsin(\frac{1}{x}) ≤ x$

$\lim_{x \to 0} x = \lim_{x \to 0} -x = 0.$ Then, by the Squeeze Theorem, $\lim_{x \to 0} x·sin(\frac{1}{x}) = 0$.

- $\lim_{x \to 0} x^{2} sin(\frac{1}{x}) = 0$ (Figure ii).

$-1 ≤ sin(\frac{1}{x}) ≤ 1 ⇨ -x^{2} ≤ x^{2}sin(\frac{1}{x}) ≤ x^{2}$

$\lim_{x \to 0} x^{2} = \lim_{x \to 0} -x^{2} = 0.$ Then, by the Squeeze Theorem, $\lim_{x \to 0} x^{2} sin(\frac{1}{x}) = 0$.

- $\lim_{x \to ∞} \frac{cos^2(2x)}{3-2x}$

From basic trigonometry, -1 ≤ cos(2x) ≤ 1 ⇒[x^{2} ≥ 0] 0 ≤ cos^{2}(2x) ≤ 1 ⇒[Since we are calculating the limit as x goes to infinity, it is completely reasonable to assume that 3 - 2x < 0] $\frac{0}{3-2x}≥\frac{cos^2(2x)}{3-2x}≥\frac{1}{3-2x}⇒ \frac{1}{3-2x}≤\frac{cos^2(2x)}{3-2x}≤0$

$\lim_{x \to ∞} 0 = 0, \lim_{x \to ∞} \frac{1}{3-2x} = 0 ⇒ \lim_{x \to ∞} \frac{cos^2(2x)}{3-2x} = 0.$

- $\lim_{x \to ∞} \frac{x^2(sin(x)+cos^3(x))}{(x^2+1)(x-3)}$

From basic trigonometry, -1 ≤ cos(x) ≤ 1, -1 ≤ sin(x) ≤ 1 ⇒ -1 ≤ cos^{3}(x) ≤ 1 ⇒ -2 ≤ sin(x) + cos^{3}(x) ≤ 2 ⇒ [Since we are calculating the limit as x goes to minus infinity, it is completely reasonable to assume that x-3 < 0] $\frac{2}{x-3}≤\frac{sin(x)+cos^3(x)}{x-3}≤\frac{-2}{x-3}$

Next, we are going to multiply by x^{2} and divide by x^{2} +1 (both positive),

$\frac{2x^2}{(x^2+1)(x-3)}≤\frac{x^2(sin(x)+cos^3(x))}{(x^2+1)(x-3)}≤\frac{-2x^2}{(x^2+1)(x-3)}$

$\lim_{x \to ∞} \frac{2x^2}{(x^2+1)(x-3)} = \lim_{x \to ∞}\frac{2x^2}{x^3-3x^2+x-3}$ =[Apply L’Hôpital’s rule or divide by x^{3} 🏮] = $\lim_{x \to ∞}\frac{\frac{2}{x}}{1-3\frac{1}{x}+\frac{1}{x^2}-3\frac{1}{x^3}} = 0$. Mutatis mutandis, $\lim_{x \to ∞} \frac{-2x^2}{(x^2+1)(x-3)}$ = 0, and so by the squeezing theorem, $\lim_{x \to ∞} \frac{x^2(sin(x)+cos^3(x))}{(x^2+1)(x-3)} = 0.$

🏮 To evaluate the limits at infinity for a rational function, we divide the numerator and denominator by the highest power of x appearing in the denominator. This determines which term in the overall expression dominates the behavior of the function at large values of x.

- $\lim_{x \to 0} (\frac{sin(x)}{x}) = 1,$ (Figure iii)

Let’s prove it geometrically. The arc length of a circle in radians can be expressed as, Arc Length = $\theta·r = x~ because~ r=1.$ Notice that as x→0, the bow string is the same size than the bow itself, that is, $\lim_{x \to 0} (\frac{sin(x)}{x}) = 1.$ -1.b.-

Another way of proving it is as follows: -1.a.-

$Area(\triangle ADF) ≥ Area(Sector ADB) ≥ Area(\triangle ADB)$
$ ⇨ \frac{tan(x)}{2} ≥ \frac{x}{2} ≥ \frac{sin(x)}{2}$ because the Area of a circle with an angle measuring 360º is πr^{2} ⇨ Area of a sector (θ is measured in degrees) is ^{θ}⁄_{360º}πr^{2} ⇨ Area(Sector ADB) = ^{1}⁄_{2}r^{2}θ, where θ is the angle in radians =[r = 1] ^{1}⁄_{2}θ, and then consider that x = θ.

$ ⇨ \frac{sin(x)}{cos(x)} ≥ x ≥ sin(x) ⇨ \frac{cos(x)}{sin(x)} ≤ \frac{1}{x} ≤\frac{1}{sin(x)} ⇨ cos(x) ≤ \frac{sin(x)}{x} ≤1$

$\lim_{x \to 0} cos(x) = \lim_{x \to 0} 1 = 1$ ⇨[by the Squeeze Theorem] $\lim_{x \to 0} (\frac{sin(x)}{x}) = 1.$

- $\lim_{x \to 0} (\frac{1-cos(x)}{x}) = 0,$ (Figure iv).

$\lim_{x \to 0} cos(\frac{\pi-x}{2}) = \lim_{x \to 0} 0 = 0, \lim_{x \to 0} 0 = 0$ ⇨[By the squeeze theorem] $\lim_{x \to 0} (\frac{1-cos(x)}{x}) = 0.$

It can also be solved as follows,

$\lim_{x \to 0} (\frac{1-cos(x)}{x}) = \lim_{x \to 0} (\frac{(1-cos(x))(1+cos(x))}{x(1+cos(x))}) = \lim_{x \to 0} (\frac{1-cos^{2}(x)}{x(1+cos(x))}) = \lim_{x \to 0} (\frac{sin^{2}(x)}{x(1+cos(x))})$

$ = \lim_{x \to 0} (\frac{sin(x)}{x}\frac{sin(x)}{1+cos(x)}) = \lim_{x \to 0} (\frac{sin(x)}{x}) \lim_{x \to 0}(\frac{sin(x)}{1+cos(x)})$ [by the product law for limits] = 1·0 = 0.

- $\lim_{x \to 0} x^2·e^{sin(\frac{1}{x})}$

$-1 ≤ sin(\frac{1}{x}) ≤ 1 ⇒$[The exponential is a monotone increasing function] $e^{-1} ≤ e^{sin(\frac{1}{x})} ≤ e^1 ⇒ x^2e^{-1} ≤ x^2e^{sin(\frac{1}{x})} ≤ x^2e$

$\lim_{x \to 0} x^2e = e·\lim_{x \to 0} x^2 = e·0 = 0, \lim_{x \to 0} x^2e^{-1} = e^{-1}·\lim_{x \to 0} x^2 = e^{-1}·0 = 0$ ⇒[By the Squeeze Theorem] $\lim_{x \to 0} x^2·e^{sin(\frac{1}{x})} = 0.$

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