Question

Задание приложено...

Answer 1

Ответ:

1)

$\boxed{ \boldsymbol{ \displaystyle \lim_{x \to 3} \frac{x - 3}{\sqrt{x - 2} - 1 } =2} }$

2)

$\boxed{ \boldsymbol{ \displaystyle \lim_{x \to -2} \frac{2 + x}{1 - \sqrt{x^{2} -3} } =0,5 } }$

3)

$\boxed{ \boldsymbol{ \displaystyle \lim_{x \to 3} \frac{\sqrt{x + 13} - 2\sqrt{x + 1} }{x^{2} - 9} = -\dfrac{1}{16} } }$

4)

$\boxed{ \boldsymbol{ \displaystyle \lim_{x \to 5} \frac{\sqrt{6 - x} - 1 }{3 - \sqrt{4 +x} } = 3 } }$

5)

$\boxed{ \boldsymbol{ \displaystyle \lim_{x \to 0} \frac{\sqrt{x^{2} + 1} -1 }{\sqrt{x^{2} + 16} -4} = 4 } }$

6)

$\boxed{ \boldsymbol{ \displaystyle \lim_{x \to 0} \frac{x}{\sqrt[3]{x + 1} - 1 } =3 } }$

Примечание:

$\lim_{x \to a} f(x) = f(a)$ если $\exists f(a)$

Объяснения:

1)

$\displaystyle \lim_{x \to 3} \frac{x - 3}{\sqrt{x - 2} - 1 } = \lim_{x \to 3} \frac{(x - 3)(\sqrt{x - 2} + 1)}{(\sqrt{x - 2} - 1)(\sqrt{x - 2} + 1) } = \lim_{x \to 3} \frac{(x - 3)(\sqrt{x - 2} + 1)}{(\sqrt{x - 2})^{2} - 1^{2} }=$

$\displaystyle = \lim_{x \to 3} \frac{(x - 3)(\sqrt{x - 2} + 1)}{(x - 3) }=\lim_{x \to 3}\sqrt{x - 2} + 1 = \sqrt{3 - 2} + 1 = \sqrt{1} + 1 = 1 + 1 =2$

2)

$\displaystyle \lim_{x \to -2} \frac{2 + x}{1 - \sqrt{x^{2} -3} } = \lim_{x \to -2} \frac{(2 + x)(1 + \sqrt{x^{2} -3})}{(1 - \sqrt{x^{2} -3})(1 + \sqrt{x^{2} -3}) } =$

$=\displaystyle \lim_{x \to -2} \frac{(2 + x)(1 + \sqrt{x^{2} -3})}{1^{2} -(\sqrt{x^{2} - 3})^{2} } = \lim_{x \to -2} \frac{(2 + x)(1 + \sqrt{x^{2} -3})}{1 - (x^{2} -3) } = \lim_{x \to -2} \frac{(2 + x)(1 + \sqrt{x^{2} -3})}{1 - x^{2} +3 }=$

$= \displaystyle \lim_{x \to -2} \frac{(2 + x)(1 + \sqrt{x^{2} -3})}{4 - x^{2} }=\lim_{x \to -2} \frac{(2 + x)(1 + \sqrt{x^{2} -3})}{(2 - x)(2 + x) } = \lim_{x \to -2} \frac{(1 + \sqrt{x^{2} -3})}{(2 - x) } =$

$= \dfrac{(1 + \sqrt{(-2)^{2} -3})}{(2 - (-2)) } = \dfrac{1 + \sqrt{4 - 3} }{2 + 2} = \dfrac{1 + \sqrt{1} }{4} = \dfrac{2}{4} = 0,5$

3)

$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 13} - 2\sqrt{x + 1} }{x^{2} - 9} = \lim_{x \to 3} \frac{(\sqrt{x + 13} - 2\sqrt{x + 1})(\sqrt{x + 13} + 2\sqrt{x + 1} ) }{(x - 3)(x + 3)(\sqrt{x + 13} + 2\sqrt{x + 1} )} =$

$=\displaystyle \lim_{x \to 3} \frac{(\sqrt{x + 13})^{2} - (2\sqrt{x + 1})^{2} }{(x - 3)(x + 3)(\sqrt{x + 13} + 2\sqrt{x + 1} )} = \lim_{x \to 3} \frac{x + 13 - 4(x +1) }{(x - 3)(x + 3)(\sqrt{x + 13} + 2\sqrt{x + 1} )} =$

$\displaystyle= \lim_{x \to 3} \frac{x + 13 - (4x +4) }{(x - 3)(x + 3)(\sqrt{x + 13} + 2\sqrt{x + 1} )}=$

$\displaystyle = \lim_{x \to 3} \frac{(x + 13 - 4x - 4) }{(x - 3)(x + 3)(\sqrt{x + 13} + 2\sqrt{x + 1} )} = \lim_{x \to 3} \frac{9 - 3x }{(x - 3)(x + 3)(\sqrt{x + 13} + 2\sqrt{x + 1} )} =$

$\displaystyle = \lim_{x \to 3} \frac{9 - 3x}{(x - 3)(x + 3)(\sqrt{x + 13} + 2\sqrt{x + 1} )}= \lim_{x \to 3} \frac{3(3 - x)}{(x - 3)(x + 3)(\sqrt{x + 13} + 2\sqrt{x + 1} )}=$

$\displaystyle = \lim_{x \to 3} \frac{-3(x - 3)}{(x - 3)(x + 3)(\sqrt{x + 13} + 2\sqrt{x + 1} )}= \lim_{x \to 3} \frac{-3}{(x + 3)(\sqrt{x + 13} + 2\sqrt{x + 1} )}=$

$\displaystyle = \lim_{x \to 3} \frac{-3}{(3 + 3)(\sqrt{3 + 13} + 2\sqrt{3 + 1} )}= -\frac{3}{6(\sqrt{16} + 2\sqrt{4} )} = -\frac{1}{2(4 + 2 \cdot 2)} =$

$= - \dfrac{1}{2(4 + 4) } = - \dfrac{1}{2 \cdot 8 } = -\dfrac{1}{16}$

4)

$\displaystyle \lim_{x \to 5} \frac{\sqrt{6 - x} - 1 }{3 - \sqrt{4 +x} } = \lim_{x \to 5} \frac{(\sqrt{6 - x} - 1)(\sqrt{6 - x} + 1)(3 + \sqrt{4 +x}) }{(3 - \sqrt{4 +x})(3 + \sqrt{4 +x})(\sqrt{6 - x} + 1) }=$

$\displaystyle = \lim_{x \to 5} \frac{((\sqrt{6 - x})^{2} - 1^{2})(3 + \sqrt{4 +x}) }{(3^{2} - (\sqrt{4 +x})^{2} )(\sqrt{6 - x} + 1)}=\lim_{x \to 5} \frac{(6 - x - 1)(3 + \sqrt{4 +x}) }{(9 - (4 + x))(\sqrt{6 - x} + 1) }=$

$=\displaystyle \lim_{x \to 5} \frac{(5 - x)(3 + \sqrt{4 +x}) }{(9 - 4 - x)(\sqrt{6 - x} + 1) }=\lim_{x \to 5} \frac{(5 - x)(3 + \sqrt{4 +x}) }{(5 - x)(\sqrt{6 - x} + 1) } =$

$\displaystyle = \lim_{x \to 5} \frac{(3 + \sqrt{4 +x}) }{(\sqrt{6 - x} + 1) } =\frac{3 + \sqrt{4 +5} }{\sqrt{6 - 5} + 1 } = \frac{3 + \sqrt{9} }{\sqrt{1} + 1 } = \frac{3 + 3}{1 + 1} = \frac{6}{2} = 3$

Примеры 5) и 6) смотрите на фотографии!!!