Limits

To be a star, you must shine your own light, follow your path, and don’t worry about the darkness, for that is when the stars shine brightest. Always do what you are afraid to do, Ralph Waldo.

Recall

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

• The pair (x, y) is denoted as y = f(x) where: x is the independent variable (input) and y is the dependent variable (output). Often, both the domain D and codomain E are the set of real numbers ℝ or subsets of ℝ. A mathematical function can be thought of as a black box (or machine) that takes an input from its domain and produces exactly one output in its codomain. Inside the machine lives a specific rule (formula, procedure, or mapping) that dictates or tells you which output corresponds to each input, and a key property is uniqueness —each input maps to a single, deterministic output. No input can ever produce two different results (Figure E). The function f(x) = x² accepts any real number x (domain: ℝ) and returns exactly one non-negative value x² (output in codomain: [0,∞)). The input 3 always yields 9, never any other value.

• D is the domain, the set of all possible inputs. E is the codomain or range, the set of all possible outputs.

• Key property💡: Each input has exactly one output. (No input is assigned two different outputs — this is the vertical line test!)

• Examples: constant, f(x) = c, horizontal line, slope = 0; linear, f(x) = mx + b, straight line, constant slope m and y-intercept b; quadratic $f(x) = ax^2 + bx + c$, u-shaped or inverted U, opens up (a > 0) or down (a < 0), vertex at x = $\frac{-b}{2a}$, symmetry about vertical axis through vertex; polynomial, $f(x) = a_n x^n + \dots + a_0$, a smooth and continuous curve, n roots (counting multiplicity), end behaviour determined by its leading term $a_n x^n$; exponential function, $f(x) = a \cdot b^x, a \ne 0, b \gt 0$, rapid growth (b > 1) or decay (0 < b < 1); trigonometric functions, $\sin(x), \cos(x), \tan (z)$ oscillatory, periodic behavior (period 2π for sin/cos, π for tan), sin and cos are bounded between -1 and 1, but tan is unbounded; step function $f(x) = \lfloor x \rfloor$, greatest integer ≤ x, constant on intervals [n, n+1), jumps at integers, its graph is a staircase shape; absolute value f(x) = |x|, V-shaped graph, slope changes at 0.

• Functions can be expressed in multiple forms, each useful in different contexts: verbal description, table of values (list of pairs), algebraic formula, graph, piecewise definition, recursive definition, parametric or integral form, and series representation.

• Evaluating a function means finding or computing the output value f(x) for a given input value x. f(x) = $x^2-2x +4, f(2) = 2^2 -2\cdot 2 + 4 = 4 - 4 + 4 = 4, f(0) = 0^2 -2\cdot 0 + 4 = 4, f(1) = 1^2 -2\cdot 1 + 4 = 1 -2 +4 = 3$

• The x-intercept is any point on the graph that intersects or crosses the x-axis. In other words, it is the value of x when the function (y-coordinate or y-value) is zero. The y-intercept is the point where the graph intersects or crosses the y-axis. y-coordinate of the point whose x-coordinate is 0, e.g., 2x - 3y = 6. x-intercept: set y = 0 → 2x = 6 ⇒ x=3, so (3, 0). y-intercept: set x = 0 → −3y = 6 ⇒ y = −2, so (0, −2).

• f is said to have a local or relative maximum at c if there exists an interval (a, b) containing c ($c \in (a, b)$) such that f(c) ≥ f(x) $∀ x \in (a, b) ∩ D$. f is said to have a local or relative minimum at c if there exists an interval (a, b) containing c ($c \in (a, b)$) such that f(c) ≤ f(x) $∀ x \in (a, b) ∩ D$.💡Local extrema can only occur where the function stops rising or falling —either because the derivative is zero, the derivative doesn't exist, or you're at the edge of the domain.

• First Derivative Test. Let f be differentiable on an open interval containing c, except possibly at c itself, and let c be a critical point (so f′(c) = 0 or f′ is undefined). Then:

1. If f′(x) > 0 for x just to the left of c and f′(x) < 0 for x just to the right of c, then f(c) is a local maximum.
2. If f′(x) < 0 for x just to the left of c and f′(x) > 0 for x just to the right of c, then f(c) is a local minimum.
3. If f′(x) does not change sign at c (stays positive or stays negative), then f(c) is not a local extremum.

• Second Derivative Test. If the first derivate of a function at a point c is zero (f'(c) = 0) and the second derivate at this point exists, then: (i) If f''(c) > 0, then f(c) is a local minimum. (ii) If f''(c) < 0, then f(c) is a local maximum. (iii) If f''(c) = 0, the test is inconclusive.
• Vertical asymptotes are vertical lines (perpendicular to the x-axis) of the form x = a (where a is a constant) near which the function grows without bound when x approaches a from at least one side. The line x = a is a vertical asymptote of f if at least one of the one-sided limits is infinite: $\lim_{x \to a^{-}}f(x)=\pm\infty$ or $\lim_{x \to a^{+}}f(x)=\pm\infty$.
• Horizontal asymptotes are horizontal lines (y = c, parallel to the x-axis) that the graph of the function approaches as x grows very large (positively or negatively). In other words, the function values get arbitrarily close to c as long as x is sufficiently large or more formally: $\lim_{x \to \infty}f(x)=c$ and/or $\lim_{x \to -\infty}f(x)=c$.
• A line y = m+b is an oblique (slant) asymptote as $x \to \infty$ if $\lim_{x\to\infty} [f(x) - (mx+b)] = 0.$ This means that the curve of f(x) gets more and more closer to the line y = mx + b as x grows large. For an asymptote y = mx + b as $x \to \infty, m = \lim_{x\to\infty} \frac{f(x)}{x}, b = \lim_{x\to\infty} \bigl(f(x) - m x\bigr)$, provided both limits exist.

Intuitive definition

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.

A limit involves what is going on around a given point, say “a”. It is the value that a function approaches as its input approaches that point “a” but does not care what the function is at this particular point, f(a).

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 is read as “the limit of f as x approaches a equals L”. This means that the value of the function f can be made arbitrarily close to L (and I mean as close as you want, e.g., L ± 0.1, L ± 0.01, L ± 0.001, you get the idea 😄), by choosing values of x sufficiently close to a.

Let’s consider f(x) = 2x² - 3x + 1 and calculate $\lim_{x \to 2} 2x^2 -3x +1$. As x approaches 2, f(x) approaches 2·(2)² - 3·(2) + 1 = 2·4 -3·2 + 1 = 3.

If you evaluate the function x² at x = 0, the function equals 0, and $\lim_{x \to 0} x^2 = 0$ (Figure 1.b). Similarly, $\lim_{x \to 2} 5x -4 = 5·2 -4 = 6$ (Figure 1.c).

• Consider the function $g(x) = \frac{x^2-1}{x-1}$, g(x) is a line, namely x + 1, but the function is undefined at x = 1. Dom(g) = ℝ - {1}. However, $\lim_{x \to 1} \frac{x^2-1}{x-1} = \lim_{x \to 1} \frac{(x-1)(x+1)}{x-1} = \lim_{x \to 1} x +1 = 2.$ To be honest, the function is begging to be redefined as f(x) = x + 1, and then Dom(f) = ℝ.

• $f(x) = \begin{cases} 3x + 1, &x ≠ 2\\\\\\\\ 2, &x < 0 \end{cases}$

$\lim_{x \to -2} f(x) = -5.$

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

Approaching 0 from the left (x < 0), f(x) becomes -2x + 2 and we get closer and closer to 2. Approaching 0 from the right (x > 0), f(x) becomes x + 1 and we get closer and closer to 1.

Therefore, $\lim_{x \to 1} f(x)$ does not exist because the left and right limits are different.

How To Evaluate Limits From a Graph

Consider f(x) = $\frac{sin(x)}{x}$ (1.h) and observe its graphical representation. Examine the graph as x gets closer and closer to the specified value. If the function approaches a specific value as x gets closer to the point, that value is the limit, e.g., $\lim_{x \to 0} \frac{sin(x)}{x} = 1$

If the function approaches positive or negative infinity, the limit is +∞ or -∞, respectively, e.g., $\lim_{x \to a} f(x) = +∞$ (Figure 1.d), $\lim_{x \to 0⁺} \frac{1}{x} = +∞$, $\lim_{x \to 0⁻} \frac{1}{x} = -∞$ (Figure 1.f). The right limit ($\lim_{x \to 0⁺} \frac{1}{x}$) means you approach the point x = 0 from the right, i.e., with values bigger than 0. The left limit ($\lim_{x \to 0⁻} \frac{1}{x}$) means you approach the point x = 0 from the right, i.e., we must take values lower than 0, hence $\lim_{x \to 0} \frac{1}{x}$ is undefined.

If the function oscillates or does not approach a specific value, the limit does not exist, e.g., $\lim_{x \to a} sin(\frac{1}{x})$ D.N.E. (does not exist, Figure 1.i).

Precise definition

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 -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\$  

Solved easy exercises

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

  Finding a limit by factoring is a technique to computing limits that works by canceling out common factors. It allows us to transform an indeterminate form into one that allows for direct evaluation. Notice that the limit as x approaches -2 has nothing no to do with the value of the function at x = -2.

• $\lim_{x \to 2}\frac{x^2-4}{x-2} =\lim_{x \to 2}\frac{(x+2)(x-2)}{x-2} = \lim_{x \to 2}(x+2) = 4$
• $\lim_{x \to 4} x^{2} + x -11 = 9.$
• $\lim_{x \to 4} \frac {x + 3}{x^{2} + 12} = 0.25.$
• $\lim_{x \to 4} \frac{\sqrt{x}-2}{x-4} = \lim_{x \to 4} \frac{(\sqrt{x}-2)(\sqrt{x}+2)}{(x-4)(\sqrt{x}+2)} = \lim_{x \to 4} \frac{x-4}{(x-4)(\sqrt{x}+2)} = \lim_{x \to 4} \frac{1}{\sqrt{x}+2} = \frac{1}{4}.$
• $\lim_{x \to 0} \frac{x}{3-\sqrt{x+9}} = \lim_{x \to 0} \frac{x(3+\sqrt{x+9})}{(3-\sqrt{x+9})(3+\sqrt{x+9})} = \lim_{x \to 0} \frac{x(3+\sqrt{x+9})}{9-x+9} = \lim_{x \to 0} \frac{x(3+\sqrt{x+9})}{-x} = \lim_{x \to 0} -(3+\sqrt{x+9}) = -(3+3) = -6.$
• $\lim_{x \to -2} \frac{x^2+5x+6}{x^2+6x+8} = \lim_{x \to -2} \frac{(x+2)(x+3)}{(x+2)(x+4)} = \lim_{x \to -2} \frac{(x+3)}{(x+4)}=\frac{1}{2}.$
• $\lim_{x \to 0}\frac{|x|}{x}, \lim_{x \to 0^-}\frac{|x|}{x} = \lim_{x \to 0^-}\frac{-x}{x} = -1, \lim_{x \to 0^+}\frac{|x|}{x} = \lim_{x \to 0^+}\frac{x}{x} = 1$. Since the one-sided limits exist but are not equal (-1 ≠ 1), the limit $\lim_{x \to 0}\frac{|x|}{x}$ do not exist. This is the canonical example of a function with a jump discontinuity at x = 0.

  A limit $\lim_{x \to a} f(x)$ exists iff both one-sided limits exist and are equal.

Solved examples with epsilon-delta proofs

• $\lim_{x \to 0} x^{2} = 0$ -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 $|x^{2}|=|x|^{2}<(\sqrt \epsilon)^{2} = \epsilon$

• $\lim_{x \to 2} 5x - 4 = 6$ -1.c.-

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

$\forall \epsilon>0, \exists \delta>0: 0<|x-2|<\delta,~implies~ 5|x - 2|<\epsilon$ because |5x -4 -6| = |5x-10| = |5·(x-2)| = |5||x-2| = 5|x-2|.

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

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

• $\lim_{x \to 4} (2x -3) = 5$

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

$|2x - 3 -5|<\epsilon ↭ 2|x-4|$ ⇒[Let δ = ε/2] $|2x - 3 -5| = 2|x-4| < 2\frac{\epsilon}{2} = \epsilon$

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

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

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

We know that $0<|x-5|<\delta$. Let’s choose δ < 1, |x-5| < δ ↭ -1 < x -5 < 1 ↭[Let’s add 2] 1 < x -3 < 3 ⇒ $\frac{1}{3} < \frac{1}{x-3} < 1 ⇒ -1 < \frac{1}{3} < \frac{1}{x-3} < 1 ⇒ -1 < \frac{1}{x-3} < 1 ⇒ |\frac{1}{x-3}| < 1 ⇒ \frac{1}{|x-3|} < 1$

[🚀] $|\frac{1}{x-3} -\frac{1}{2}| = \frac{1}{2}·|x-5|·\frac{1}{|x-3|} < \frac{1}{2}·δ·1$ [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 $0<|x|<\delta,~implies~ |xsin(\frac {1}{x})| = |x|·|sin(\frac {1}{x})| ≤ |x|< δ = ε$∎

• $\lim_{x \to 2} 3x^2 = 12$

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

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

Let’s choose δ = min{1, something else -be patience😄-}, |x-2| < 1 ↭ -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] $\frac{3|x-5|}{\sqrt{3x+1} + 4}$ [$\sqrt{3x+1} + 4>1 ⇒ \frac{1}{\sqrt{3x+1} + 4} < \frac{1}{1}$] 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] $\frac{1}{2}< x-2 < \frac{3}{2} ⇒ 2 > \frac{1}{x-2} > \frac{2}{3} > -2 ⇒ \frac{1}{|x-2|}< 2$

[🚀] $|\frac{x}{x-2} - 3|$ < 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·(|x²|+|2x|+|4|) < δ(3² +2·3 + 4) = δ·19 [Choose δ = min{1, ε/19}] ε ∎

Bibliography

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