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1a,$\mathop {\lim }\limits_{x \to -1}$ $\dfrac{x^{2} - 3}{x^{3} + 2}=\dfrac{(-1)^{2} - 3}{(-1)^{3} + 2}=\dfrac{-2}{1}=-2$
1a,.$\mathop {\lim }\limits_{x \to -1}$ $\dfrac{x^{2} - 3}{x^{3} + 2}=\dfrac{(-1)^{2} - 3}{(-1)^{3} + 2}=\dfrac{-2}{1}=-2$
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1b.$\mathop {\lim }\limits_{x \to 3}\left(\dfrac{4x-3}{2x+7}\right)^5=\left(\dfrac{4.3-3}{2.3+7}\right)^5=\dfrac{9^5}{13^5}$
1b..$\mathop {\lim }\limits_{x \to 3}\left(\dfrac{4x-3}{2x+7}\right)^5=\left(\dfrac{4.3-3}{2.3+7}\right)^5=\dfrac{9^5}{13^5}$
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1c.$\mathop {\lim }\limits_{x \to -2}\sqrt[3]{\dfrac{2x^4+3x+2}{x^2-x+2}}=\sqrt[3]{\dfrac{2.(-2)^4+3.(-2)+2}{(-2)^2-(-2)+2}}=\sqrt[3]{\dfrac{7}{2}}$
1c..$\mathop {\lim }\limits_{x \to -2}\sqrt[3]{\dfrac{2x^4+3x+2}{x^2-x+2}}=\sqrt[3]{\dfrac{2.(-2)^4+3.(-2)+2}{(-2)^2-(-2)+2}}=\sqrt[3]{\dfrac{7}{2}}$
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1d.$\mathop {\lim }\limits_{x \to 3}\sqrt{\dfrac{x^2}{x^3-x-6}}=\sqrt{\dfrac{3^2}{3^3-3-6}}=\dfrac{1}{\sqrt2}$
1d..$\mathop {\lim }\limits_{x \to 3}\sqrt{\dfrac{x^2}{x^3-x-6}}=\sqrt{\dfrac{3^2}{3^3-3-6}}=\dfrac{1}{\sqrt2}$
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1e.$\mathop {\lim }\limits_{x \to 1}\sqrt{\dfrac{5x-1}{2x+7}}=\sqrt{\dfrac{5.1-1}{2.1+7}}=\dfrac{2}{3}$
1e.$\mathop {\lim }\limits_{x \to 1}\sqrt{\dfrac{5x-1}{2x+7}}=\sqrt{\dfrac{5.1-1}{2.1+7}}=\dfrac{2}{3}.$
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1f.$\mathop {\lim }\limits_{x \to -2}\dfrac{x^2+5x+3}{2x^3+2x^2+x+6}=\dfrac{(-2)^2+5.(-2)+3}{2.(-2)^3+2.(-2)^2+(-2)+6}=\dfrac{3}{4}$
1f.$\mathop {\lim }\limits_{x \to -2}\dfrac{x^2+5x+3}{2x^3+2x^2+x+6}=\dfrac{(-2)^2+5.(-2)+3}{2.(-2)^3+2.(-2)^2+(-2)+6}=\dfrac{3}{4}$.
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Bài 2:1. $\mathop {\lim }\limits_{x \to 2}\dfrac{x^2+3x-10}{3x^2-5x-2}$$=\mathop {\lim }\limits_{x \to 2}\dfrac{(x-2)(x+5)}{(x-2)(3x+1)}$$=\mathop {\lim }\limits_{x \to 2}\dfrac{x+5}{3x+1}=1$
Bài 2:1. $\mathop {\lim }\limits_{x \to 2}\dfrac{x^2+3x-10}{3x^2-5x-2}$$=\mathop {\lim }\limits_{x \to 2}\dfrac{(x-2)(x+5)}{(x-2)(3x+1)}$$=\mathop {\lim }\limits_{x \to 2}\dfrac{x+5}{3x+1}=1.$
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Bài 2:2. $\mathop {\lim }\limits_{x \to a}\dfrac{x^n-a^n}{x-a}$$=\mathop {\lim }\limits_{x \to a}\dfrac{(x-a)(x^{n-1}+x^{n-2}a+\ldots+xa^{n-2}+a^{n-1})}{x-a}$$=\mathop {\lim }\limits_{x \to a}(x^{n-1}+x^{n-2}a+\ldots+xa^{n-2}+a^{n-1})=na^{n-1}$
Bài 2:2.. $\mathop {\lim }\limits_{x \to a}\dfrac{x^n-a^n}{x-a}$$=\mathop {\lim }\limits_{x \to a}\dfrac{(x-a)(x^{n-1}+x^{n-2}a+\ldots+xa^{n-2}+a^{n-1})}{x-a}$$=\mathop {\lim }\limits_{x \to a}(x^{n-1}+x^{n-2}a+\ldots+xa^{n-2}+a^{n-1})=na^{n-1}$
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Câu 2)10, $\mathop {\lim }\limits_{x \to 1}\dfrac{x^{4} - 1}{x^{2} + 2x - 3}=\mathop {\lim }\limits_{x \to 1}\dfrac{(x-1)(x+1)(x^2+1)}{(x-1)(x+3)}=\mathop {\lim }\limits_{x \to 1}\dfrac{(x+1)(x^2+1)}{x+3}=1$
Câu 2)10, $\mathop {\lim }\limits_{x \to 1}\dfrac{x^{4} - 1}{x^{2} + 2x - 3}=\mathop {\lim }\limits_{x \to 1}\dfrac{(x-1)(x+1)(x^2+1)}{(x-1)(x+3)}=\mathop {\lim }\limits_{x \to 1}\dfrac{(x+1)(x^2+1)}{x+3}=1.$
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Câu 29, $\mathop {\lim }\limits_{x \to 1}\dfrac{x^{3} - 1}{x(x+ 5) - 6}=\mathop {\lim }\limits_{x \to 1}\dfrac{(x-1)(x^2+x+1)}{(x-1)(x+6)}=\mathop {\lim }\limits_{x \to 1}\dfrac{(x^2+x+1)}{(x+6)}=\dfrac{3}{7}$
Câu 29, $\mathop {\lim }\limits_{x \to 1}\dfrac{x^{3} - 1}{x(x+ 5) - 6}=\mathop {\lim }\limits_{x \to 1}\dfrac{(x-1)(x^2+x+1)}{(x-1)(x+6)}=\mathop {\lim }\limits_{x \to 1}\dfrac{(x^2+x+1)}{(x+6)}=\dfrac{3}{7}.$
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Câu 2) 8, $\mathop {\lim }\limits_{x \to - 5}\dfrac{x^{2} + 2x - 15}{x + 5}=\mathop {\lim }\limits_{x \to - 5}\dfrac{(x+5)(x-3)}{x + 5}=\mathop {\lim }\limits_{x \to - 5}(x-3)=-8$
Câu 28, $\mathop {\lim }\limits_{x \to - 5}\dfrac{x^{2} + 2x - 15}{x + 5}=\mathop {\lim }\limits_{x \to - 5}\dfrac{(x+5)(x-3)}{x + 5}=\mathop {\lim }\limits_{x \to - 5}(x-3)=-8$
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Câu 27, $\mathop {\lim }\limits_{x \to 3} \dfrac{x^{2} + 2x -1}{x - 3}=\mathop {\lim }\limits_{x \to 3} \dfrac{1}{x - 3}.\mathop {\lim }\limits_{x \to 3}(x^{2} + 2x -1)=(\infty).(3^{2} + 6 -1)=+\infty$
Câu 2.7, $\mathop {\lim }\limits_{x \to 3} \dfrac{x^{2} + 2x -1}{x - 3}=\mathop {\lim }\limits_{x \to 3} \dfrac{1}{x - 3}.\mathop {\lim }\limits_{x \to 3}(x^{2} + 2x -1)=(\infty).(3^{2} + 6 -1)=+\infty$
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Câu 2.6, $\mathop {\lim }\limits_{x \to 1}\dfrac{x - 1}{1 - \sqrt{x}}=\mathop {\lim }\limits_{x \to 1}\dfrac{x - 1}{\dfrac{1-x}{1 + \sqrt{x}}}=-\mathop {\lim }\limits_{x \to 1}\dfrac{1}{1 + \sqrt{x}}=-\dfrac{1}{2}$
Câu 2..6, $\mathop {\lim }\limits_{x \to 1}\dfrac{x - 1}{1 - \sqrt{x}}=\mathop {\lim }\limits_{x \to 1}\dfrac{x - 1}{\dfrac{1-x}{1 + \sqrt{x}}}=-\mathop {\lim }\limits_{x \to 1}\dfrac{1}{1 + \sqrt{x}}=-\dfrac{1}{2}$
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Bài 2:3. $\mathop {\lim }\limits_{x \to a}\dfrac{x^n-a^n-na^{n-1}(x-a)}{(x-a)^2}$$=\mathop {\lim }\limits_{x \to a}\dfrac{(x-a)(x^{n-1}+x^{n-2}a+\ldots+xa^{n-2}+a^{n-1}-na^{n-1})}{(x-a)^2}$$=\mathop {\lim }\limits_{x \to a}\dfrac{(x^{n-1}-a^{n-1})+a(x^{n-2}-a^{n-2})+\ldots+a^{n-3}(x^2-a^2)+a^{n-2}(x-a)}{x-a}$$=\mathop {\lim }\limits_{x \to a}\dfrac{(x-a)(x^{n-2}+x^{n-3}a+\ldots+xa^{n-3}+a^{n-2})+a(x-a)(x^{n-3}+x^{n-4}a+\ldots+xa^{n-4}+a^{n-3})+\ldots+a^{n-3}(x-a)(x+a)+a^{n-2}(x-a)}{x-a}$$=\mathop {\lim }\limits_{x \to a}[(x^{n-2}+x^{n-3}a+\ldots+xa^{n-3}+a^{n-2})+a(x^{n-3}+x^{n-4}a+\ldots+xa^{n-4}+a^{n-3})+\ldots+a^{n-3}(x+a)+a^{n-2}]$$=\dfrac{n(n-1)}{2}a^{n-2}$
Bài 2:3.. $\mathop {\lim }\limits_{x \to a}\dfrac{x^n-a^n-na^{n-1}(x-a)}{(x-a)^2}$$=\mathop {\lim }\limits_{x \to a}\dfrac{(x-a)(x^{n-1}+x^{n-2}a+\ldots+xa^{n-2}+a^{n-1}-na^{n-1})}{(x-a)^2}$$=\mathop {\lim }\limits_{x \to a}\dfrac{(x^{n-1}-a^{n-1})+a(x^{n-2}-a^{n-2})+\ldots+a^{n-3}(x^2-a^2)+a^{n-2}(x-a)}{x-a}$$=\mathop {\lim }\limits_{x \to a}\dfrac{(x-a)(x^{n-2}+x^{n-3}a+\ldots+xa^{n-3}+a^{n-2})+a(x-a)(x^{n-3}+x^{n-4}a+\ldots+xa^{n-4}+a^{n-3})+\ldots+a^{n-3}(x-a)(x+a)+a^{n-2}(x-a)}{x-a}$$=\mathop {\lim }\limits_{x \to a}[(x^{n-2}+x^{n-3}a+\ldots+xa^{n-3}+a^{n-2})+a(x^{n-3}+x^{n-4}a+\ldots+xa^{n-4}+a^{n-3})+\ldots+a^{n-3}(x+a)+a^{n-2}]$$=\dfrac{n(n-1)}{2}a^{n-2}$
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Bài 24.Áp dụng phần 3 với $a=1$ ta có:$\mathop {\lim }\limits_{x \to 1}\dfrac{x^n-nx+n-1}{(x-1)^2}=\dfrac{n(n-1)}{2}$
Bài 24.Áp dụng phần 3 với $a=1$ ta có:$\mathop {\lim }\limits_{x \to 1}\dfrac{x^n-nx+n-1}{(x-1)^2}=\dfrac{n(n-1)}{2}.$
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