Úlohy - limita - úlohy s odmocninami
Úloha
\(\lim_{x \to 1} \frac {x^2 - x}{\sqrt{x} - 1} =\)
- \(= \lim_{x \to 1} \frac {(x^2 - x)(\sqrt{x} + 1)}{(\sqrt{x} - 1)(\sqrt{x} + 1)} = \)
- \(= \lim_{x \to 1} \frac {x(x - 1)(\sqrt{x} + 1)}{(x - 1)} = \)
- \(= \lim_{x \to 1} x(\sqrt{x} + 1) = \)
- \(= 2\)
Úloha
\(\lim_{x \to 0} \frac {1 - \sqrt{1 - x}}{x} =\)
- \(= \lim_{x \to 0} \frac {(1 - \sqrt{1 - x})(1 + \sqrt{1 - x})}{x(1 + \sqrt{1 - x})} = \)
- \(= \lim_{x \to 0} \frac {1 - (1 - x)}{x(1 + \sqrt{1 - x})} = \)
- \(= \lim_{x \to 0} \frac {x}{x(1 + \sqrt{1 - x})} = \)
- \(= \lim_{x \to 0} \frac {1}{1(1 + \sqrt{1 - x})} = \)
- \(= \frac {1}{2}\)
Úloha
\(\lim_{x \to 3} \frac {9 - x^2}{\sqrt{3x} - 3} =\)
- \(= \lim_{x \to 3} \frac {(9 - x^2)(\sqrt{3x} + 3)}{(\sqrt{3x} - 3)(\sqrt{3x} + 3)} = \)
- \(= \lim_{x \to 3} \frac {(3 - x)(3 + x)(\sqrt{3x} + 3)}{(3x - 9)} = \)
- \(= \lim_{x \to 3} \frac {-(x - 3)(3 + x)(\sqrt{3x} + 3)}{3(x - 3)} = \)
- \(= \lim_{x \to 3} \frac {-(3 + x)(\sqrt{3x} + 3)}{3} = \)
- \(= -12\)
Úloha
\(\lim_{x \to -3} \frac {x + 3}{\sqrt{x + 4} - 1} =\)
- \(= \lim_{x \to -3} \frac {x + 3}{\sqrt{x + 4} - 1} \cdot \frac{\sqrt {x + 4} + 1}{\sqrt{x + 4} + 1} = \)
- \(= \lim_{x \to -3} \frac {(x + 3)(\sqrt{x + 4} + 1)}{x + 4 - 1} = \)
- \(= \lim_{x \to -3} (\sqrt{x + 4} + 1) = \)
- \(= 2\)
Úloha
\(\lim_{x \to 7} \frac {2 - \sqrt{x - 3}}{x^2 - 49} =\)
- \(= \lim_{x \to 7} \frac {2 - \sqrt{x - 3}}{x^2 - 49} \cdot \frac{2 + \sqrt{x - 3}}{2 + \sqrt{x - 3}} = \)
- \(= \lim_{x \to 7} \frac {4 - (x - 3)}{(x^2 - 49)(2 + \sqrt{x - 3})} = \)
- \(= \lim_{x \to 7} \frac {7 - x}{(x - 7)(x + 7)(2 + \sqrt{x - 3})} = \)
- \(= \lim_{x \to 7} \frac {-1}{(x + 7)(2 + \sqrt{x - 3})} = \)
- \(= -\frac {1}{56}\)
Úloha
\(\lim_{x \to 0} \frac {x}{\sqrt{1 + x} - \sqrt{1 - x}} =\)
- \(= \lim_{x \to 0} \frac {x}{\sqrt{1 + x} - \sqrt{1 - x}} \cdot \frac {\sqrt{1 + x} + \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}} = \)
- \(= \lim_{x \to 0} \frac {x({\sqrt{1 + x} + \sqrt{1 - x}})}{1 + x - (1 - x)} = \)
- \(= \lim_{x \to 0} \frac {x(\sqrt{1 + x} + \sqrt{1 - x})}{2x} = \)
- \(= \lim_{x \to 0} \frac {\sqrt{1 + x} + \sqrt{1 - x}}{2} = \)
- \(= 1\)
Úloha
\(\lim_{x \to a} \frac {a - x}{\sqrt{a} - \sqrt{x}} =\)
- \(= \lim_{x \to a} \frac {a - x}{\sqrt{a} - \sqrt{x}} \cdot \frac{\sqrt{a} + \sqrt{x}}{\sqrt{a} + \sqrt{x}} = \)
- \(= \lim_{x \to a} \frac {(a - x)(\sqrt{a} + \sqrt{x})}{a - x} = \)
- \(= \lim_{x \to a} (\sqrt{a} + \sqrt{x}) = \)
- \(= 2\sqrt{a}\)
