Question

Задание приложено...

Answer 1

Ответ:

1)

$\boxed{ \boldsymbol{ \displaystyle \lim_{x \to 1} \dfrac{x^{2} - 3x + 2}{x^{2} - 4x + 3} = 0,5 } }$

2)

$\boxed{ \boldsymbol{ \displaystyle \lim_{x \to 1} \dfrac{x^{4} - 3x + 2}{x^{5} - 4x + 3} =1 } }$

3)

$\boxed{ \boldsymbol{ \displaystyle \lim_{x \to 0} \dfrac{(1 + x)^{4} - (1 + 4x)}{x^{2} + x^{4}} = 6 } }$

4)

$\boxed{ \boldsymbol{ \lim_{x \to 1} \dfrac{x^{5} - 1}{x^{3} - 1} = \frac{5}{3} } }$

Примечание:

$\lim_{x \to a} f(a) = f(a)$ если  $f$ существует в точке $a$

------------------------------------------------------------------------------------------------------

Если:

1) $\lim_{x \to a} f(x) = \lim_{x \to a} g(x) = 0$

2) функции $f(x)$ и $g(x)$ дифференцируемы в окрестности точки $x = a$

3) $\exists \lim_{x \to a} \dfrac{f'(x)}{g'(x)}$

По правилу Лопиталя:

$\boxed{ \lim_{x \to a} \dfrac{f(x)}{g(x)} = \bigg [ \dfrac{0}{0} \bigg ] = \lim_{x \to a} \dfrac{f'(x)}{g'(x)} }$

Объяснение:

1)

$\displaystyle \lim_{x \to 1} \dfrac{x^{2} - 3x + 2}{x^{2} - 4x + 3} = \bigg [\frac{0}{0} \bigg ] = \lim_{x \to 1} \dfrac{(x^{2} - 3x + 2)'}{(x^{2} - 4x + 3)'} = \lim_{x \to 1} \dfrac{2x - 3}{2x - 4} = \dfrac{2 \cdot 1 - 3}{2 \cdot 1 - 4}=$

$= \dfrac{2 - 3}{2 - 4}= \dfrac{-1}{-2} = 0,5$

2)

$\displaystyle \lim_{x \to 1} \dfrac{x^{4} - 3x + 2}{x^{5} - 4x + 3} = \bigg [\frac{0}{0} \bigg ] =\lim_{x \to 1} \dfrac{(x^{4} - 3x + 2)'}{(x^{5} - 4x + 3)'} = \lim_{x \to 1} \dfrac{4x^{3} - 3}{5x^{4} - 4} = \dfrac{4 \cdot 1^{3} - 3}{5 \cdot 1^{4} - 4} =$

$= \dfrac{4 - 3}{5 - 4} = \dfrac{1}{1} = 1$

3)

$\displaystyle \lim_{x \to 0} \dfrac{(1 + x)^{4} - (1 + 4x)}{x^{2} + x^{4}} = \displaystyle \lim_{x \to 0} \dfrac{(1 + x)^{4} - 4x - 1}{x^{2} + x^{4}} = \bigg [\dfrac{0}{0} \bigg ] = \lim_{x \to 0} \dfrac{((1 + x)^{4} - 4x - 1)'}{(x^{2} + x^{4})'} =$

$= \lim_{x \to 0} \dfrac{4(1 + x)^{3}(x + 1)' - 4}{2x + 4x^{3}} = \lim_{x \to 0} \dfrac{4(1 + x)^{3} - 4}{2x + 4x^{3}} = \bigg [\dfrac{0}{0} \bigg ] =$

$= \lim_{x \to 0} \dfrac{(4(1 + x)^{3} - 4)'}{(2x + 4x^{3})'} = \lim_{x \to 0} \dfrac{12(1 + x)^{2} }{2 + 12x^{2}} = \dfrac{12(1 + 0)^{2} }{2 + 12 \cdot 0^{2}} = \dfrac{12}{2} = 6$

4)

$\lim_{x \to 1} \dfrac{x^{5} - 1}{x^{3} - 1} = \bigg [\dfrac{0}{0} \bigg] = \lim_{x \to 1} \dfrac{(x^{5} - 1)'}{(x^{3} - 1)'} = \lim_{x \to 1} \dfrac{5x^{4}}{3x^{2} } = \lim_{x \to 1} \dfrac{5x^{2}}{3 } =$

$= \dfrac{5 \cdot 1^{2}}{3 } = \dfrac{5}{3}$