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\(\lim\limits_{x\rightarrow3^+}\frac{7x-1}{x-3}=\frac{20}{0}=+\infty\)
\(\lim\limits_{x\rightarrow5^+}\frac{11-2x}{x-5}=\frac{1}{0}=+\infty\)
\(\lim\limits_{x\rightarrow3^-}\frac{-x-3}{3-x}=\frac{-6}{0}=-\infty\)
\(a=\lim\limits_{x\rightarrow1}\frac{\left(\sqrt{3x+1}-\sqrt{x+3}\right)\left(\sqrt{3x+1}+\sqrt{x+3}\right)}{\left(x-1\right)\left(x+1\right)\left(\sqrt{3x+1}+\sqrt{x+3}\right)}=\lim\limits_{x\rightarrow1}\frac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)\left(\sqrt{3x+1}+\sqrt{x+3}\right)}\)
\(=\lim\limits_{x\rightarrow1}\frac{2}{\left(x+1\right)\left(\sqrt{3x+1}+\sqrt{x+3}\right)}=\frac{2}{2.4}=\frac{1}{4}\)
\(b=\frac{3}{0}=+\infty\)
\(c=\frac{-13}{0}=-\infty\)
Lời giải:
a)
\(\lim\limits_{x\to-1}\frac{\sqrt[3]{x}+1}{2x^2+5x+3}=\lim\limits_{x\to-1}\frac{x+1}{\left(\sqrt[3]{x^2}-\sqrt[3]{x}+1\right)\left(x+1\right)\left(2x+3\right)}\)
\(\lim\limits_{x\to-1}\frac{1}{\left(\sqrt[3]{x^2}-\sqrt[3]{x}+1\right)\left(2x+3\right)}=\frac{1}{\left(\sqrt[3]{\left(-1\right)^2}-\sqrt[3]{-1}+1\right)\left(2.-1+3\right)}=\frac{1}{3}\)
b)
\(\lim\limits_{x\to1}\frac{\sqrt[3]{x^2}-2\sqrt[3]{x}+1}{\left(x-1\right)^2}=\lim\limits_{x\to1}\frac{\left(\sqrt[3]{x}-1\right)^2}{\left(x-1\right)^2}=\lim\limits_{x\to1}\frac{\left(x-1\right)^2}{\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)^2\left(x-1\right)^2}\)
\(=\lim\limits_{x\to1}\frac{1}{\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)^2}=\frac{1}{\left(1+1+1\right)^2}=\frac{1}{9}\)
c)
\(\lim_{x\to 1}\frac{\sqrt[4]{x}-1}{x^3+x^2-2}=\lim_{x\to 1}\frac{\sqrt[4]{x}-1}{(x-1)(x^2+2x+2)}=\lim_{x\to 1}\frac{x-1}{(\sqrt{x}+1)(\sqrt[4]{x}+1)(x-1)(x^2+2x+2)}\)
\(=\lim_{x\to 1}\frac{1}{(\sqrt{x}+1)(\sqrt[4]{x}+1)(x^2+2x+2)}=\frac{1}{(1+1)(1+1)(1+2.1+2)}=\frac{1}{20}\)
d)
\(\lim_{x\to -2}\frac{\sqrt[3]{2x+12}+x}{x^2+2x}=\lim_{x\to -2}\frac{2x+12+x^3}{(\sqrt[3]{(2x+12)^2}-x\sqrt[3]{2x+12}+x^2).x(x+2)}\)
\(=\lim_{x\to -2}\frac{(x+2)(x^2-2x+6)}{(\sqrt[3]{(2x+12)^2}-x\sqrt[3]{2x+12}+x^2).x(x+2)}=\lim_{x\to -2}\frac{x^2-2x+6}{(\sqrt[3]{(2x+12)^2}-x\sqrt[3]{2x+12}+x^2).x}\)
\(=\frac{-7}{12}\)
Lời giải:
a)
\(\lim\limits_{x\to-1}\frac{\sqrt[3]{x}+1}{2x^2+5x+3}=\lim\limits_{x\to-1}\frac{x+1}{\left(\sqrt[3]{x^2}-\sqrt[3]{x}+1\right)\left(x+1\right)\left(2x+3\right)}\)
\(\lim\limits_{x\to-1}\frac{1}{\left(\sqrt[3]{x^2}-\sqrt[3]{x}+1\right)\left(2x+3\right)}=\frac{1}{\left(\sqrt[3]{\left(-1\right)^2}-\sqrt[3]{-1}+1\right)\left(2.-1+3\right)}=\frac{1}{3}\)
b)
\(\lim\limits_{x\to1}\frac{\sqrt[3]{x^2}-2\sqrt[3]{x}+1}{\left(x-1\right)^2}=\lim\limits_{x\to1}\frac{\left(\sqrt[3]{x}-1\right)^2}{\left(x-1\right)^2}=\lim\limits_{x\to1}\frac{\left(x-1\right)^2}{\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)^2\left(x-1\right)^2}\)
\(=\lim\limits_{x\to1}\frac{1}{\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)^2}=\frac{1}{\left(1+1+1\right)^2}=\frac{1}{9}\)
c)
\(\lim_{x\to 1}\frac{\sqrt[4]{x}-1}{x^3+x^2-2}=\lim_{x\to 1}\frac{\sqrt[4]{x}-1}{(x-1)(x^2+2x+2)}=\lim_{x\to 1}\frac{x-1}{(\sqrt{x}+1)(\sqrt[4]{x}+1)(x-1)(x^2+2x+2)}\)
\(=\lim_{x\to 1}\frac{1}{(\sqrt{x}+1)(\sqrt[4]{x}+1)(x^2+2x+2)}=\frac{1}{(1+1)(1+1)(1+2.1+2)}=\frac{1}{20}\)
d)
\(\lim_{x\to -2}\frac{\sqrt[3]{2x+12}+x}{x^2+2x}=\lim_{x\to -2}\frac{2x+12+x^3}{(\sqrt[3]{(2x+12)^2}-x\sqrt[3]{2x+12}+x^2).x(x+2)}\)
\(=\lim_{x\to -2}\frac{(x+2)(x^2-2x+6)}{(\sqrt[3]{(2x+12)^2}-x\sqrt[3]{2x+12}+x^2).x(x+2)}=\lim_{x\to -2}\frac{x^2-2x+6}{(\sqrt[3]{(2x+12)^2}-x\sqrt[3]{2x+12}+x^2).x}\)
\(=\frac{-7}{12}\)
Liên hợp thì thật khủng khiếp, tạm thời xử lý bằng L'Hopital:
\(\lim\limits_{x\rightarrow1}\frac{\left(2x-1\right)^{\frac{1}{4}}+\left(x-2\right)^{\frac{1}{5}}}{x-1}=\lim\limits_{x\rightarrow1}\frac{\frac{1}{2}\left(2x-1\right)^{-\frac{3}{4}}+\frac{1}{5}\left(x-2\right)^{-\frac{4}{5}}}{1}=\frac{1}{2}+\frac{1}{5}=\frac{7}{10}\)
Mình nghĩ bạn bị sai đề:
Bạn thử sửa đề lại thành:
lim (x--> 2) \(\frac{\sqrt{2x+5}-\sqrt{7+x}}{x^2-2x}\)
\(_{x\underrightarrow{lim}2}\frac{\sqrt{2x+5}-\sqrt{7-x}}{x^2-2x}\)
\(=x\underrightarrow{lim}2\frac{\left(\sqrt{2x+5}-\sqrt{7+x}\right)\left(\sqrt{2x+5}+\sqrt{7+x}\right)}{\left(x^2-2x\right)\left(\sqrt{2x+5}+\sqrt{7+x}\right)}\)
\(=x\underrightarrow{lim}2\frac{1}{x\left(\sqrt{2x+5}+\sqrt{7+x}\right)}=\frac{1}{12}\)
\(\lim\limits_{x\rightarrow1}\frac{\sqrt{x}-1}{x-1}=\lim\limits_{x\rightarrow1}\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\lim\limits_{x\rightarrow1}\frac{1}{\sqrt{x}+1}=\frac{1}{2}\)
\(\lim\limits_{x\rightarrow2}\frac{\sqrt{4x+1}-3}{x^2-4}=\lim\limits_{x\rightarrow2}\frac{4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)\left(\sqrt{4x+1}+3\right)}=\lim\limits_{x\rightarrow2}\frac{4}{\left(x+2\right)\left(\sqrt{4x+1}+3\right)}=\frac{1}{6}\)
Lời giải:
\(\lim\limits_{x\to-1}\frac{x+\sqrt{4x+5}}{\sqrt{7-2x}-\sqrt{x+10}}=\lim\limits_{x\to-1}\frac{x^2-\left(4x+5\right)}{x-\sqrt{4x+5}}.\frac{\sqrt{7-2x}+\sqrt{x+10}}{7-2x-\left(x+10\right)}\)
\(=\lim\limits_{x\to-1}\frac{\left(x-5\right)\left(x+1\right)}{x-\sqrt{4x+5}}.\frac{\sqrt{7-2x}+\sqrt{x+10}}{-3\left(x+1\right)}\)
\(=\lim\limits_{x\to-1}\frac{\left(x-5\right)\left(\sqrt{7-2x}+\sqrt{x+10}\right)}{-3\left(x-\sqrt{4x+5}\right)}=\frac{\left(-1-5\right)\left(\sqrt{7-2.-1}+\sqrt{-1+10}\right)}{-3\left(-1-\sqrt{4.-1+5}\right)}=-6\)
Bậc tử lớn hơn bậc mẫu, giới hạn bằng \(+\infty\)
Cụ thể thì:
\(=lim\frac{\sqrt{x+\frac{1}{x^3}-\frac{11}{x^4}}}{2+\frac{1}{x}+\frac{1}{x^2}}=\frac{+\infty}{2}=+\infty\)