Answer to Question #131416 in Calculus for Moel Tariburu

Question

Prove using Rigorous definition of two-sided limits
lim┬(x→1)⁡〖28/(4x+3)〗=4

Answer 1

=Prove;

=lim (28/(4x + 3)) = 4
x-1

Using rigorous, we use an ℇ - ᶳ proof

Preliminary analysis: We need | f(x) – L| in terms of |x - c|

= |28/(4x + 3)(– 4) | = | 28/4 (x + 3/4) |

= |7(x + 3/4) |

=7|x + 3/4 |

Therefore, we can see that |28/(4x + 3) (– 4) | = 7|x + 3/4 |

We therefore take ᶳ = ℇ/7

Proceeding with formal proof; Let ℇ > 0 be given, ᶳ = ℇ/7 and assume that 0 < | x – 0| < ᶳ

So; =|28/(4x + 3) (– 4) | = | 28/4 (x + 3/4) |

= |7(x + 3/4) |

=7|x + 3/4 |

< 7ᶳ = 7 * ℇ/7 = ℇ

Thus, we conclude that,   |28/(4x + 3)(– 4) | < ℇ