Помогите с математикой, Решите ПРЕДЕЛЫ!
[tex] \lim_{x \to \ 7} = \frac{x^{3}-49 }{x-7}= [/tex]
[tex] \lim_{x \to \ 2} = \frac{x^{2}-5x6 }{x-2}= [/tex]
[tex] \lim_{x \to \infty} \frac{ x^{3}+4 x^{4} }{2+ x^{4} } [/tex]
[tex] \lim_{x \to \ 7} = \frac{x^{3}-49 }{x-7}= [/tex]
[tex] \lim_{x \to \ 2} = \frac{x^{2}-5x6 }{x-2}= [/tex]
[tex] \lim_{x \to \infty} \frac{ x^{3}+4 x^{4} }{2+ x^{4} } [/tex]
More Questions From This User See All
Answers & Comments
= limx---->7 [(x - 7)(x + 7)] / ( x - 7) = limx---->7 (x + 7) = (7 + 7 ) = 14
2)
= limx----->2 [(x² - 5x + 6)/(x - 2) = = limx----->2 [(x - 2)(x - 3)] /(x - 2) =
= limx----->2 (x - 3) = (2 - 3) = - 1
3) =
= limx--->∞ (x³/x⁴ + 4x⁴/x⁴) /([2/x⁴ + x⁴/x) =
= limx------( [1/x + 4) /(2/x⁴ + ) = (0 + 4)/(0 + 1) = 4/1 = 4