1.2 Limits Analyticallyap Calculus

Theorem1.3.1Basic Limit Properties
  1. 1.2 Limits Analyticallyap Calculus Frq
  2. 1.2 Limits Analyticallyap Calculus 14th Edition

Let (btext{,}) (ctext{,}) (L) and (K) be real numbers, let (n) be a positive integer, and let (f) and (g) be functions with the following limits: begin{align*}lim_{xto c}f(x)amp=Lamplim_{xto c} g(x)amp = Ktext{.}end{align*}

Given a function y = f. ( x) and an x -value, c, we say that “the limit of the function f, as x approaches c, is a value L ”: 1. If “ y tends to L ” as “ x tends to c.”. If “ y approaches L ” as “ x approaches c.”. If “ y is near L ” whenever “ x is near c.”. Limits describe how a function behaves near a point, instead of at that point. This simple yet powerful idea is the basis of all of calculus. If you're seeing this message, it means we're having trouble loading external resources on our website. 1.2 Limits Analytically Name: Write your questions and thoughts here! Calculus Recall: What is a limit? Notes Finding a limit: 1. Direct Substitution Factor and Cancel 1. 5: T 62 T F4; 2. Lim → 6 √3 F2 3. Lim → 4 4 65 T T 4. Lim →?; 2 6 E13 F7 T E7 Rationalize Two variables 5.

1.2 Limits Analyticallyap Calculus Frq

The following limits hold.

Constant

(limlimits_{xto c} b = b)

Identity
Limits

(limlimits_{xto c} x = c)

Sum/Difference

(limlimits_{xto c}(f(x)pm g(x)) = Lpm K)

Scalar Multiple

(limlimits_{xto c}(bcdot f(x)) = bL)

1.2 Limits Analyticallyap Calculus
Product

(limlimits_{xto c} (f(x)cdot g(x)) = LK)

Quotient

(limlimits_{xto c} (f(x)/g(x)) = L/Ktext{,}) when (Kneq 0)

Power

(limlimits_{xto c} f(x)^n = L^n)

Root

(limlimits_{xto c} sqrt[n]{f(x)} = sqrt[n]{L})

(If (n) is even, (L) must be non-negative.)

Composition

If either of the following holds:

1.2
  1. (limlimits_{xto c}f(x)=Ltext{,}) (limlimits_{xto L}g(x)=Ktext{,}) and (g(L)=K)

  2. (limlimits_{xto c}f(x)=Ltext{,}) (limlimits_{xto L}g(x)=Ktext{,}) and (f(x)neq L) for all (x) close to but not equal to (c)

Limits

then (limlimits_{xto c}g(f(x)) = Ktext{.})

Answer

$limlimits f(x)_{x to c}$ exists for all values of c except for at $c=4$

Work Step by Step

1.2 limits analyticallyap calculus 14th edition

1.2 Limits Analyticallyap Calculus 14th Edition

A limit is the $y$ value of a function as it gets infinitely close to an $x$ value. A limit exists if $lim limits f(x)_{xto c^{-}}$ $=$ $lim limits f(x)_{xto c^{+}}$. When finding limits on a piecewise function, you first check to make sure each individual 'piece' exists at the given bounds. Then you make sure that at the endpoints of each boundary, the two functions equal each other. In this problem, $limlimits f(x)_{a to 2^{-}}=4$ and $limlimits f(x)_{a to 2^{+}}=4$. However, at the other boundary $x=4$, $lim limits f(x)_{xto 4^{-}}= 0$ and $lim limits f(x)_{xto 4^{+}}=4$, so the limit does not exist at $x=4$