A limit is the value to which a function grows close as the input approaches some value. One would say that the limit of f, as x approaches a, is 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 to a* (x ∈ [a-δ, a+δ], x≠a)) -Fig 1.a.-

A limit involves what is going on around a point, and does not care what happens at it.

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

$\forall \epsilon>0, \exists \delta>0: |f(x)-L|<\epsilon, whenever~ 0<|x-a|<\delta$ Exercise: $\lim_{x \to 0} x^{2} = 0$ -1.b.-

$\forall \epsilon>0, \exists \delta>0: 0<|x|<\delta, implies~ |x^{2}|<\epsilon$

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 $|x^{2}|=|x|^{2}<(\sqrt \epsilon)^{2} = \epsilon$

Exercise: $\lim_{x \to 2} 5x - 4 = 6$ -1.c.-

$\forall \epsilon>0, \exists \delta>0: 0<|x-2|<\epsilon,~implies~ |5x - 4 -6|<\epsilon$

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

Let’s choose $\delta = \frac {\epsilon}{5}$, then

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

Exercises:

- $\lim_{x \to 4} x^{2} + x -11 = 9.$
- $\lim_{x \to 4} \frac {x + 3}{x^{2} + 12} = 0.25.$
- $\lim_{x \to 0} xsin(\frac {1}{x}) = 0.$

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

Let’s choose $\delta = \epsilon, 0<|x|<\epsilon,~implies~ |xsin(\frac {1}{x})|≤|x|<\epsilon$

Definitions. A right hand limit is defined as the value L to which a function grows close as the input approaches some value from the right.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

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

Definitions. A left hand limit is defined as the value L to which a function grows close as the input approaches some value from the left.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

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

Example: $\lim_{x \to 0^{+}} \sqrt x = 0$

$\forall \epsilon>0, \exists \delta>0: |\sqrt x|<\epsilon, whenever~ 0 < x < \delta.$

Let’s choose $\delta = \epsilon^{2}$

$\forall \epsilon>0, \exists \delta>0: 0 < x < \epsilon^{2}, then~ |\sqrt x| < \sqrt{\epsilon^{2}} = \epsilon.$

Exercise: $f(x) = \begin{cases} x + 1, &x > 0\\\\ -2x + 2, &x < 0 \end{cases}$ -1.g.-

$\lim_{x \to 0^{+}} f(x) = \lim_{x \to 0^{+}} x + 1 = 1.$ $\lim_{x \to 0^{-}} f(x) = \lim_{x \to 0^{-}} -2x + 2 = 2.$ We don’t need a value for x = 0.

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

$\forall M>0, \exists \delta>0: f(x) > M, whenever~ 0<|x-a|<\delta.$ We can make the value of f(x) arbitrary large by taking x to be sufficiently close, but not equal to a. -1.d.-

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

$\forall M<0, \exists \delta>0: f(x) < M, whenever~ 0<|x-a|<\delta.$ We can make the value of f(x) arbitrary small by taking x to be sufficiently close, but not equal to a. Examples:

- $\lim_{x \to 0} \frac {1}{x^{2}} = \infty$

$\forall M>0, \exists \delta>0: \frac {1}{x^{2}} > M, whenever~ 0<|x|<\delta.$

Let’s choose $\delta = \frac {1}{\sqrt M}$

$\forall M>0, \exists \delta = \frac {1}{\sqrt M}>0: \frac {1}{x^{2}} > M, whenever~ 0<|x|<\frac {1}{\sqrt M}.$

$|x|<\frac {1}{\sqrt M} ⇨ |x|^{2}<\frac {1}{M} ⇨ x^{2}<\frac {1}{M} ⇨ M <\frac{1}{x^{2}}$

Limits at infinity are used to describe the behavior of functions as the independent variable increases or decreases without bound.

Definition. Let f(x) be a function defined on (K, ∞) for some K. Then, we say that, $\lim_{x \to \infty} f(x) = L$, and f(x) is said to have **a horizontal asymptote at y = L** if

$\forall \epsilon>0, \exists M>0: |f(x)-L|<\epsilon, whenever~ x>M.$ We can make f(x) as close as we want to L by making x large enough. -1.e.-

Let f(x) be a function defined on (-∞, K) for some K. Then, we say that, $\lim_{x \to -\infty} f(x) = L$, and f(x) is said to have **a horizontal asymptote at y = L** if

$\forall \epsilon>0, \exists M<0: |f(x)-L|<\epsilon, whenever~ x < M$ We can make f(x) as close as we want to L by making x small enough.

Exercise: $\lim_{x \to \infty} \frac{1}{x} = 0$ -1.f.-

$\forall \epsilon>0, \exists M>0: |\frac{1}{x}|<\epsilon, whenever~ x>M$

Let’s choose $M = \frac{1}{\epsilon}, x>\frac{1}{\epsilon} ⇨ |x|>\frac{1}{\epsilon} ⇨ \epsilon>\frac{1}{|x|} = |\frac{1}{x}|$

Exercise: $\lim_{x \to \infty} \frac {cos(x)}{x} = \infty$. It intersects its horizontal asymptote (y = 0) an infinite number of times as it oscillates around it with ever-decreasing amplitude.

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. 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, $\lim_{x \to a} g(x) = L.$

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

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

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

Exercises:

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

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

$\lim_{x \to 0} x^{2} = \lim_{x \to 0} -x^{2} = 0.$

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

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 is ^{θ}⁄_{360º}πr^{2} ⇨ Area(Sector ADB) = ^{1}⁄_{2}r^{2}θ, where θ is the angle in radians = ^{1}⁄_{2}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$

$\lim_{x \to 0} cos(x) = \lim_{x \to 0} 1 = 1 ⇨ \lim_{x \to 0} (\frac{sin(x)}{x}) = 1.$

- $\lim_{x \to 0} (\frac{1-cos(x)}{x}) = 0.$ $\lim_{x \to 0} cos(\frac{\pi-x}{2}) = \lim_{x \to 0} 0 = 0 ⇨ \lim_{x \to 0} (\frac{1-cos(x)}{x}) = 0.$

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.

Let f(x), g(x) be functions defined on an interval that contains x = a, **except possibly at x = a**. assume that L and M are real numbers such that $ \lim_{x \to a} f(x) = L$ and $ \lim_{x \to a} g(x) = M$. Let c be a constant. Then, each of the following statements holds:

**Sum law for limits**. It states that the limit of the sum of two functions equals the sum of the limits of two functions, that is, $\lim_{x \to a} (f(x)+g(x)) = lim_{x \to a} f(x) + lim_{x \to a} g(x) = L + M.$

Proof: Let $\epsilon>0$

$\exists \delta_1>0: 0<|x-a|<\delta_1, implies~ |f(x)-L|<\frac{epsilon}{2}$

$\exists \delta_2>0: 0<|x-a|<\delta_2, implies~ |g(x)-M|<\frac{epsilon}{2}$

Let’s choose $\delta = min (\delta_1, \delta_2).$

$\forall \epsilon>0, \exists \delta>0: 0<|x-a|<\delta, implies~ |f(x)+g(x)-L-M| ≤ |f(x)-L|+ |g(x)-M|$ [by the triangle inequality] $<\epsilon$

**Difference law for limits**. It states that the limit of the difference of two functions equals the difference of the limits of two functions, that is, $\lim_{x \to a} (f(x)-g(x)) = lim_{x \to a} f(x) - lim_{x \to a} g(x) = L - M.$**Constant multiple law for limits**. $\lim_{x \to a} c·f(x) = c·lim_{x \to a} f(x) = c·L.$**Product law for limits**. It states that the limit of a product of functions equals the product of the limits of each function, that is, $\lim_{x \to a} (f(x)·g(x)) = lim_{x \to a} f(x) · lim_{x \to a} g(x) = L · M.$**Quotient law for limits**. $\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)} = \frac{L}{M}.$**Power law for limits**. It states that the limit of the nth power of a function equals the nth power of the limit of the function, that is, $\lim_{x \to a} (f(x))^{n} = (\lim_{x \to a} f(x))^{n} = L^{n}.$**Root law for limits**. $\lim_{x \to a} \sqrt[n]{f(x)} = \sqrt[n]{\lim_{x \to a} f(x)} = \sqrt[n]{L} $ for all L if n is odd and for L≥0 if n is even.