Problem | Solution |

\(\displaystyle \lim_{x \to 3} 2x = 6\) | rootmath |

\(\displaystyle \lim_{x \to 2} 2x + 1 = 5\) | Dave Salwinski |

\(\displaystyle \lim_{x \to a} c = c\) | Dave Salwinski |

\(\displaystyle \lim_{x \to 3} x^2 = 9\) | Dave Salwinski |

\(\displaystyle \lim_{x \to 2} x^2 = 4\) | rootmath |

\(\displaystyle \lim_{x \to 3} 2x - 1 = 5\) | Prof. Redden |

\(\displaystyle \lim_{x \to 2} (x - 2)^7 \cdot \cos\left(\dfrac{2}{x - 2}\right) = 0\)

\(\displaystyle \lim_{x \to 3} (2x - 1) = 5\)

\(\displaystyle \lim_{x \to 2} 3x - 2 = 4\)

\(\displaystyle \forall a \in \mathbb{R} : \lim_{x \to a} mx + b = ma + b\)

\(\displaystyle \forall a \in \mathbb{R} : \lim_{x \to a} x^2 = a^2\)

\(\displaystyle \forall a \in \mathbb{R} : \lim_{x \to a} x^3 = a^3\)

\(\displaystyle \forall a \in \mathbb{R} - \{0\} : \lim_{x \to a} x^3 = a^3\)