Bài 108633

Question

Tính các giới hạn sau:
a)

$\mathop {\lim }\limits_{x \to 3}\frac{x^{2}-5x+6}{x^{2}-8x+15}$

b)

$\mathop {\lim }\limits_{x \to -1}\frac{x^{3}-2x-1}{x^{5}-2x-1}$

score 0 · Answer 1

$\mathop {\lim }\limits_{x \to 3}\frac {x^ {2 }-5x+6 } {x^ {2 }-8x+15 }= \mathop {\lim }\limits_{x \to 3}\frac {(x-2)(x-3) } {(x-3)(x-5) }= \mathop {\lim }\limits_{x \to 3}\frac {x-2 } {x-5 }=-\frac {1 } { 2 }$

$\mathop {\lim }\limits_{x \to -1}\frac {x^ {3 }-2x-1 } {x^ {5 }-2x-1 }= \mathop {\lim }\limits_{x \to -1}\frac {(x+1)(x^ {2 }-x-1) } {(x+1)(x^ {4 }-x^ {3 }+x^ {2 }-x-1) }$