Пределы функций непрерывного аргумента

В данных примерах x \in R
Пример 4.1
Найти предел

    \[\lim_{x\to 5} \left(\frac{x^2+6x+7}{2x-5}+3\right)=\frac{25+30+7}{5}=\frac{62}{5}=12\frac{2}{5}\]

Пример 4.2
Найти предел

    \[\lim_{x\to -\frac{5}{2}} \frac{13x+7}{2x+5}=\frac{13\cdot\frac{-5}{2}+7}{-5+5}=\frac{-25.5}{0}=-\infty\]

Пример 4.3
Найти предел

    \[\lim_{x\to 5} \frac{x^3+2x^2-35x}{x^2-3x-10}=\left[ \frac{0}{0} \right]=\lim_{x\to 5} \frac{x(x-5)(x+7)}{(x-5)(x+2)}=\]

    \[=\lim_{x\to 5} \frac{x(x+7)}{(x+2)}=\frac{5\cdot12}{7}=\frac{60}{7}=8\frac{4}{7}\]

Пример 4.4
Найти предел

    \[\lim_{x\to -2} \left(\frac{1}{2+x}-\frac{12}{8+x^3}\right)=\left[\infty-\infty\right]=\]

    \[=\lim_{x\to -2} \left(\frac{1}{2+x}-\frac{12}{(2+x)(4-2x+x^2)}\right)=\]

    \[\lim_{x\to -2} \frac{4-2x+x^2-12}{(2+x)(4-2x+x^2)}=\lim_{x\to -2} \frac{x^2-2x-8}{(2+x)(4-2x+x^2)}=\]

    \[=\lim_{x\to -2} \frac{(x+2)(x-4)}{(2+x)(4-2x+x^2)}=\lim_{x\to -2} \frac{x-4}{4-2x+x^2}=\frac{-2-4}{4+4+4}=\]

    \[-\frac{6}{12}=-\frac{1}{2}\]

ФОРМУЛА

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http://wordpress.org/support/topic/an-unexpected-http-error-occurred-during-the-api-request-on-wordpress-3?replies=37
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http://wordpress.org/extend/plugins/core-control/
http://wordpress.org/support/topic/an-unexpected-http-error-occurred-during-the-api-request-on-wordpress-3?replies=37

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