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1/ \(\lim\limits_{x\rightarrow0}\dfrac{2\sqrt{1+x}-2+2-\sqrt[3]{8-x}}{x}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{2x}{\sqrt{1+x}+1}+\dfrac{x}{4+2\sqrt[3]{8-x}+\sqrt[3]{\left(8-x\right)^2}}}{x}\)
\(=\lim\limits_{x\rightarrow0}\left(\dfrac{2}{\sqrt{1+x}+1}+\dfrac{1}{4+2\sqrt[3]{8-x}+\sqrt[3]{\left(8-x\right)^2}}\right)=\dfrac{13}{12}\)
2/ \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x+7}-\sqrt{x+3}}{x^2-3x+2}=\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x+7}-2-\left(\sqrt{x+3}-2\right)}{\left(x-1\right)\left(x-2\right)}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{x-1}{\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4}-\dfrac{x-1}{\sqrt{x+3}+2}}{\left(x-1\right)\left(x-2\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{1}{\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4}-\dfrac{1}{\sqrt{x+3}+2}}{x-2}=\dfrac{1}{6}\)
3/ \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x^2+7}-\sqrt{5-x^2}}{x^2-1}=\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x^2+7}-2+2-\sqrt{5-x^2}}{x^2-1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{\left(x^2-1\right)}{\sqrt[3]{\left(x^2+7\right)^2}+2\sqrt[3]{x^2+7}+4}+\dfrac{x^2-1}{2+\sqrt{5-x^2}}}{x^2-1}\)
\(=\lim\limits_{x\rightarrow1}\left(\dfrac{1}{\sqrt[3]{\left(x^2+7\right)^2}+2\sqrt[3]{x^2+7}+4}+\dfrac{1}{2+\sqrt{5-x^2}}\right)=\dfrac{1}{3}\)
4/ \(\lim\limits_{x\rightarrow-2}\dfrac{\sqrt{x+11}-\sqrt[3]{8x+43}}{2x^2+3x-2}=\lim\limits_{x\rightarrow-2}\dfrac{\sqrt{x+11}-3-\left(\sqrt[3]{8x+43}-3\right)}{\left(2x-1\right)\left(x+2\right)}\)
\(=\lim\limits_{x\rightarrow-2}\dfrac{\dfrac{x+2}{\sqrt{x+11}+3}-\dfrac{8\left(x+2\right)}{\sqrt[3]{\left(8x+43\right)^2}+3\sqrt[3]{8x+43}+9}}{\left(2x-1\right)\left(x+2\right)}\)
\(=\lim\limits_{x\rightarrow-2}\dfrac{\dfrac{1}{\sqrt{x+11}+3}-\dfrac{8}{\sqrt[3]{\left(8x+43\right)^2}+3\sqrt[3]{8x+43}+9}}{2x-1}=\dfrac{7}{270}\)
5/ \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt[n]{1+ax}-\sqrt[m]{1+bx}}{x}=\lim\limits_{x\rightarrow0}\dfrac{\sqrt[n]{1+ax}-1-\left(\sqrt[m]{1+bx}-1\right)}{x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{ax}{\sqrt[n]{\left(1+ax\right)^{n-1}}+\sqrt[n]{\left(1+ax\right)^{n-2}}+...+1}-\dfrac{bx}{\sqrt[m]{\left(1+bx\right)^{m-1}}+\sqrt[m]{\left(1+ax\right)^{m-2}}+...+1}}{x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{a}{\sqrt[n]{\left(1+ax\right)^{n-1}}+\sqrt[n]{\left(1+ax\right)^{n-2}}+...+1}-\dfrac{b}{\sqrt[m]{\left(1+bx\right)^{m-1}}+\sqrt[m]{\left(1+ax\right)^{m-2}}+...+1}\)
\(=\dfrac{a}{n}-\dfrac{b}{m}\)
6/ \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}.\sqrt[3]{1+6x}-1}{x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}.\sqrt[3]{1+6x}-\sqrt{1+4x}+\sqrt{1+4x}-1}{x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}.\left(\sqrt[3]{1+6x}-1\right)+\sqrt{1+4x}-1}{x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}.\dfrac{6x}{\sqrt[3]{\left(1+6x\right)^2}+\sqrt[3]{1+6x}+1}+\dfrac{4x}{\sqrt{1+4x}+1}}{x}\)
\(=\lim\limits_{x\rightarrow0}\left(\dfrac{6\sqrt{1+4x}}{\sqrt[3]{\left(1+6x\right)^2}+\sqrt[3]{1+6x}+1}+\dfrac{4}{\sqrt{1+4x}+1}\right)=4\)
\(\lim\limits_{x\rightarrow a}\frac{sin\left(\frac{x-a}{2}\right)}{\frac{x-a}{2}}.cos\left(\frac{x+a}{2}\right)=1.cos\left(\frac{a+a}{2}\right)=cosa\)
b/ \(\lim\limits_{x\rightarrow\pi}\frac{sin\frac{\pi}{2}-sin\frac{x}{2}}{\pi-x}=\lim\limits_{x\rightarrow\pi}\frac{sin\left(\frac{\pi-x}{4}\right)}{\frac{\pi-x}{4}}.\frac{cos\left(\frac{\pi+x}{4}\right)}{2}=\frac{cos\left(\frac{\pi+\pi}{4}\right)}{2}=0\)
c/ Đặt \(x-\frac{\pi}{3}=a\Rightarrow x=a+\frac{\pi}{3}\)
\(\lim\limits_{a\rightarrow0}\frac{sina}{1-2cos\left(a+\frac{\pi}{3}\right)}=\lim\limits_{a\rightarrow0}\frac{sina}{1-cosa+\sqrt{3}sina}\)
\(=\lim\limits_{a\rightarrow0}\frac{2sin\frac{a}{2}cos\frac{a}{2}}{-2sin^2\frac{a}{2}+2\sqrt{3}sin\frac{a}{2}cos\frac{a}{2}}=\lim\limits_{a\rightarrow0}\frac{cos\frac{a}{2}}{-sin\frac{a}{2}+\sqrt{3}cos\frac{a}{2}}=\frac{1}{\sqrt{3}}\)
d/Ta có: \(tana-tanb=\frac{sina}{cosa}-\frac{sinb}{cosb}=\frac{sina.cosb-cosa.sinb}{cosa.cosb}=\frac{sin\left(a-b\right)}{cosa.cosb}\)
Áp dụng:
\(\lim\limits_{x\rightarrow a}\frac{\left(tanx-tana\right)\left(tanx+tana\right)}{\frac{sin\left(x-a\right)}{cos\left(x-a\right)}}=\lim\limits_{x\rightarrow a}\frac{sin\left(x-a\right)\left(tanx+tana\right).cos\left(x-a\right)}{sin\left(x-a\right).cosx.cosa}=\lim\limits_{x\rightarrow a}\frac{\left(tanx+tana\right).cos\left(x-a\right)}{cosx.cosa}\)
\(=\frac{2tana}{cos^2a}\)
1.
\(\lim\limits_{x\to (-1)-}\frac{\sqrt{x^2-3x-4}}{1-x^2}=\lim\limits_{x\to (-1)-}\frac{\sqrt{(x+1)(x-4)}}{(1-x)(1+x)}\)
\(=\lim\limits_{x\to (-1)-}\frac{\sqrt{4-x}}{(x-1)\sqrt{-(x+1)}}=-\infty\) do:
\(\lim\limits_{x\to (-1)-}\frac{\sqrt{4-x}}{x-1}=\frac{-\sqrt{5}}{2}<0\) và \(\lim\limits_{x\to (-1)-}\frac{1}{\sqrt{-(x+1)}}=+\infty\)
2.
\(\lim\limits_{x\to 2+}\left(\frac{1}{x-2}-\frac{x+1}{\sqrt{x+2}-2}\right)=\lim\limits_{x\to 2+}\frac{1-(x+1)(\sqrt{x+2}+2)}{x-2}=-\infty\) do:
\(\lim\limits_{x\to 2+}\frac{1}{x-2}=+\infty\) và \(\lim\limits_{x\to 2+}[1-(x+1)(\sqrt{x+2}+2)]=-11<0\)
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)
\(\lim_{x\to -2}\frac{\sqrt[3]{2x+12}+x}{x^2+2x}=\lim_{x\to -2}\frac{2x+12+x^3}{(\sqrt[3]{(2x+12)^2}-x\sqrt[3]{2x+12}+x^2).x(x+2)}\)
\(=\lim_{x\to -2}\frac{(x+2)(x^2-2x+6)}{(\sqrt[3]{(2x+12)^2}-x\sqrt[3]{2x+12}+x^2).x(x+2)}=\lim_{x\to -2}\frac{x^2-2x+6}{(\sqrt[3]{(2x+12)^2}-x\sqrt[3]{2x+12}+x^2).x}\)
\(=\frac{-7}{12}\)
\(A=\lim\limits_{x\rightarrow0}\frac{\left(1+ax\right)^{\frac{1}{n}}-1}{x}=\lim\limits_{x\rightarrow0}\frac{\frac{a}{n}\left(1+ax\right)^{\frac{1-n}{n}}}{1}=\frac{a}{n}\)
\(B=\lim\limits_{x\rightarrow0}\frac{\left(1+ax\right)^{\frac{1}{n}}-1}{\left(1+bx\right)^{\frac{1}{m}}-1}=\lim\limits_{x\rightarrow0}\frac{\frac{a}{n}\left(1+ax\right)^{\frac{1-n}{n}}}{\frac{b}{m}\left(1+bx\right)^{\frac{1-m}{m}}}=\frac{am}{bn}\)
\(C=\lim\limits_{x\rightarrow0}\frac{\sqrt[3]{1+bx}\sqrt[4]{1+cx}\left(\sqrt{1+ax}-1\right)+\sqrt[4]{1+cx}\left(\sqrt[3]{1+bx}-1\right)+\left(\sqrt[4]{1+cx}-1\right)}{x}\)
\(C=\lim\limits_{x\rightarrow0}\sqrt[3]{1+bx}\sqrt[4]{1+cx}.\frac{\sqrt{1+ax}-1}{x}+\lim\limits_{x\rightarrow0}\sqrt[4]{1+cx}.\frac{\sqrt[3]{1+bx}-1}{x}+\lim\limits_{x\rightarrow0}\frac{\sqrt[4]{1+cx}-1}{x}\)
Từ câu A ta có: \(\lim\limits_{x\rightarrow0}\frac{\sqrt[n]{1+ax}-1}{x}=\frac{a}{n}\)
\(\Rightarrow C=\frac{a}{2}+\frac{b}{3}+\frac{c}{4}\)
Bạn sử dụng định lý L'Hopital cho giới hạn vô định:
\(\lim\limits_{x\rightarrow a}\frac{f\left(x\right)}{g\left(x\right)}=\lim\limits_{x\rightarrow a}\frac{f'\left(x\right)}{g'\left(x\right)}\)
\(a=\lim\limits_{x\rightarrow1^+}\frac{\sqrt{x-1}+\sqrt{x}-1}{\sqrt{\left(x-1\right)\left(x+1\right)}}=\lim\limits_{x\rightarrow1^+}\left(\frac{1}{\sqrt{x+1}}+\frac{x-1}{\left(\sqrt{x}+1\right)\sqrt{\left(x-1\right)\left(x+1\right)}}\right)\)
\(=\lim\limits_{x\rightarrow1^+}\left(\frac{1}{\sqrt{x+1}}+\frac{\sqrt{x-1}}{\left(\sqrt{x}+1\right)\sqrt{x+1}}\right)=\frac{1}{\sqrt{2}}+0=\frac{1}{\sqrt{2}}\)
\(b=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(x^{n-1}+x^{n-2}+...+x+1\right)}{\left(x-1\right)\left(x^{m-1}+x^{m-2}+...+x+1\right)}=\lim\limits_{x\rightarrow1}\frac{x^{n-1}+x^{n-2}+...+1}{x^{m-1}+x^{m-2}+...+1}=\frac{n}{m}\)
\(c=\lim\limits_{x\rightarrow1}\frac{x-1+x^2-1+...+x^n-1}{x-1}=\lim\limits_{x\rightarrow1}\frac{x-1}{x-1}+\lim\limits_{\rightarrow1}\frac{x^2-1}{x-1}+...+\lim\limits_{x\rightarrow1}\frac{x^n-1}{x-1}\)
Áp dụng kết quả câu b ta được:
\(c=\frac{1}{1}+\frac{2}{1}+...+\frac{n}{1}=1+2+..+n=\frac{n\left(n+1\right)}{2}\)
Cảm ơn bạn nhé!