Ответ:

1. 0

2. [tex]1 \frac{1}{2}[/tex]

3. [tex]\frac{1}{2}[/tex]

4. 0

5. 2

6. В 6 не видно предела, к сожалению.

Объяснение:

1. [tex]\lim_{x \to 3} \frac{x^{2}-9}{x^{2} +x-6} = \frac{3^{2} -9}{3^{2}+3-6} = 0[/tex]

2. [tex]\lim_{x \to 2} \frac{x^{2}+2x-8 }{x^{2} -4} = \lim_{x \to 2} \frac{\frac{d}{dx}(x^{2} +2x-8)}{\frac{d}{dx}(x^{2} -4) } = \lim_{x \to 2} \frac{2x+2}{2x} = \lim_{x \to 2} \frac{x+1}{x} = \frac{2+1}{2} = \frac{3}{2} =1 \frac{1}{2}[/tex]

3. [tex]\lim_{x \to 5}\frac{5x-x^{2} }{25-x^{2} } = \lim_{x \to 5}\frac{\frac{d}{dx} (5x-x^{2})}{\frac{d}{dx}(25-x^{2} ) } = \lim_{x \to 5} \frac{5-2x}{-2x} =\frac{5-2*5}{-2*5} = \frac{1}{2}[/tex]

4. [tex]\lim_{x \to -3} \frac{x^{2}+6x+9 }{x^{2} -9} = \lim_{x \to -3} \frac{\frac{d}{dx}(x^{2} +6x+9) }{\frac{d}{dx}(x^{2} -9)} = \lim_{x \to -3} \frac{2x+6}{2x} =\lim_{x \to -3} \frac{2(x+3)}{2x} =\frac{-3+3}{-3} = 0[/tex]

5.[tex]\lim_{x \to 3} \frac{x^{3}-9x-x^{2} +9 }{x^{3}- 4x^{2} +3x} = \lim_{x \to 3}\frac{\frac{d}{dx}(x^{3}-9x-x^{2} +9) }{\frac{d}{dx}(x^{3}- 4x^{2} +3x) } = \lim_{x \to 3} \frac{3x^{2} -2x-9}{3x^{2} -8x+3} = \frac{3*3^{2}-2*3-9}{3*3^{2} -8*3+3}=2[/tex]