Matematyka
xxxasiiek
2017-09-29 01:11:55
Zadanie 4 oblicz granice od podpunktu d) ,do f)
Odpowiedź
olusia240399
2017-09-29 04:22:20

[latex]\d) \ \ lim_{x o-3 } dfrac{ sqrt{x+4}-1 }{2x+6} =lim_{x o-3 } dfrac{ (sqrt{x+4}-1 )( sqrt{x+4} +1)}{2(x+3)( sqrt{x+4} +1)} = \ \ \lim_{x o-3 } dfrac{ x+4-1 }{2(x+3)( sqrt{x+4}+1) } =lim_{x o-3 } dfrac{ x+3 }{2(x+3)( sqrt{x+4}+1) } = \ \ dfrac{1}{2( sqrt{-3+4} +1)} = dfrac{1}{2cdot2} = frac{1}{4} \ \e) \ \ lim_{x o1} dfrac{(x^3-1)( sqrt{x^2+3}+2) }{(sqrt{x^2+3}-2)(sqrt{x^2+3}+2)} [/latex] [latex]\ lim_{x o1} dfrac{(x-1)(x^2+x+1)( sqrt{x^2+3}+2) }{x^2+3-4} = \ \ \lim_{x o1} dfrac{(x-1)(x^2+x+1)( sqrt{x^2+3}+2) }{(x+1)(x-1)} = \ \ \ dfrac{(1+1+1)( sqrt{1+3}+2 )}{1+1} = dfrac{3cdot4}{2} =6 \ \f) \ \ lim_{x o -2} dfrac{(x^2-x-6)(1+ sqrt{x^2-3} )}{(1- sqrt{x^2-3})(1+ sqrt{x^2-3} ) } = \ \ \ lim_{x o -2} dfrac{(x^2+2x-3x-6)(1+ sqrt{x^2-3} )}{1-x^2+3 } = \ \ lim_{x o -2} dfrac{(x^2+2x-3x-6)(1+ sqrt{x^2-3} )}{1-x^2+3 } = [/latex] [latex]\lim_{x o -2} dfrac{[x(x+2)-3(x+2)](1+ sqrt{x^2-3} )}{(2-x)(2+x) } = \ \ \lim_{x o -2} dfrac{(x+2)(x-3)(1+ sqrt{x^2-3} )}{(2-x)(2+x) } = dfrac{(-2-3)(1+1)}{2-(-2)} =- dfrac{5}{2} [/latex]

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