29/09/2014

# Lecture and Home Assignment

## Find the indicated limit.

Q:01 Lim(x Ã 0) (3x | Q:02 Lim(x Ã 2) (2x |

Q:03 Lim(x Ã 2) {(x | Q:04 Lim(x Ã 3) (2x – 8)/(x + 4) |

Q:05 Lim(x Ã 2) (2x | Q:06 Lim(x Ã -4) 250 |

Q:07 Lim(x Ã 0) 175 | Q:08 Lim(x Ã -10) (x |

Q:09 Lim(x Ã 1) (3x | Q:10 Lim(x Ã -2) (-x |

Q:11 Lim(x Ã -2) (- x | Q:12 Lim(x Ã -2) (5x |

Q:13 Lim(x Ã 5) (4x - 20)/(3x | Q:14 Lim(x Ã -2) (10 + x |

Q:15 Lim(x Ã -3) (6x | Q:16 Lim(x Ã 5) [{(x + 3)/(x + 6)}{x |

Q:17 Lim(x Ã -4) (x | Q:18 Lim(x Ã 1) (x |

Q:19 Lim(x Ã -4) (x | Q:20 Lim(x Ã 9) (81 – x |

Q:21 Lim(x Ã a) (4x | Q:22 Lim(x Ã -d) (x |

Definite state of function: example:

f(x) = x^{2 } + 2x Ã¨ = f(2) = (2)^{2} + 2(2) Ã¨ = 8 -: so 8 is a real/definite number so function is definite:-

Horizontal asymptotes: when answer is in definite number or real number.

Example: lim(x Ã ∞) f(x) then lim(x Ã ∞) (3x^{2 }+ 5x)/(4x^{2 }5) Ã¨ = ¾

Vertical asymptotes: opposite to horizontal asymptotes.

(Q: 25 Page683) Solution:

H.A V.A

lim(x Ã ∞) -3x/(5x + 100) by putting x value -20

lim(x Ã ∞) x (-3)/ x {5+ (100/x)} lim(x Ã -20) -3x/(5x + 100)

lim(x Ã ∞) -3/{5 + (100/x)} -3(-20)/{5(-20) + 100)

-3/{5 + (100/∞)} = ** -3/5** 60/(-100 +100) =

__∞__## Find the indicated limit & comment on the existence of any asymptotes.

Q:23 Lim(x Ã ∞) 4 /x | Q:24 Lim(x Ã ∞) (5x – 3)/(x + 10) |

:25 Lim(x Ã ∞) -3x/(5x+100) | Q:26 Lim(x Ã ∞) (8x + 10)/- 4x |

Q:27 Lim(x Ã ∞) 2x | Q:28 Lim(x Ã ∞) (x |

Q:29 Lim(x Ã ∞) -8x/(4x + 1000) | Q:30 Lim(x Ã ∞) (5x + 10000)/(2x – 5000) |

Q:31 Lim(x Ã ∞) (100 – 3x | Q:32 Lim(x Ã ∞) (3x |

Continuity: the continuity exists when 2 terms are validated the terms are below:

Ã˜ Function is definite at x = a at x = 2

Ã˜ Lim(x Ã a) f(x) = F(a)

1. Example of continues: f(x) = X^{3 }

Solution: f(a) = f(x) so

F(a) f(x)

Put x = 2 lim(x Ã 2) x^{3}

F(2) = (2)^{3 } = 8 answer (2)^{3 }= 8 answer

So function is continues

2. Example of discontinues No. 01: f(x) = 1 / (x^{3 }– x)

Solution: Check where function is discontinues

x^{3} – x = 0 Ã¨ = x (x^{2} – 1) = 0 Ã¨ = x (x – 1) (x + 1) = 0 Ã¨ x = 0, x = 1 , x = -1

f(x) at 0 Ã¨ = f(0) = 1/{(0)^{3} – 0} Ã¨ = 1/0 ∞ answer

f(x) at 1 Ã¨ = f(1) = 1/{(1)^{3 }– 1} Ã¨ = 1/0 ∞ answer

f(x) at -1 Ã¨ = f(-1) = 1/{(-1)^{3 }- (-1)} Ã¨ = 1/0 ∞ answer

3. Example of discontinues No. 02: f(x) = 3x^{2}/ (x + 5)

Solution: Check where function is discontinues

X + 5 = 0 Ã¨ = x = -5

F(x) at -5 Ã¨ = f(-5) = 3(-5)^{2}/{(-5) + 5} Ã¨ = 28/0 ∞ answer

## Find discontinuity

Q:33 f(x) = 3x | Q:34 f(x) = 1/(x + 2) |

Q:35 f(x) = x | Q:36 f(x) = 3x |

Q:37 f(x) = 1/(8-2x) | Q:38 f(x) = |x| |

Q:39 f(x) = (2x + 3)/(x | Q:40 f(x) = (x – 2)/(x |

Q:41 f(x) = (4x – 3)/(x | Q:42 f(x) = (5/x)/(2x |

Q:43 f(x) = 20/(x | Q44 f(x) = (4/x)/(18 + 3x – x |

Q:45 f(x) = {10/(5 – x)}/ (4 – x | Q46 f(x) = {5/(3 – x)}/(x |

Q:47 f(x) = (3x – 5)/(x | Q:48 f(x) = {3/(x |

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