Tính limx→+∞x+1x+2…x+nn−x bằng

Câu hỏi:

Tính $\underset{x\to +\infty }{\lim }\left(\sqrt[n]{\left(x+1\right)\left(x+2\right)\mathrm{\dots }\left(x+n\right)}-x\right)$ bằng

A. 0

B. $\frac{n+1}{2}$

Đáp án chính xác

C. n

D. 1

Trả lời:

Đặt $x=\frac{1}{y}$ khi $x\to +\infty :y\to 0$$\begin{array}{l}\underset{x\to +\infty }{\lim }\left(\sqrt[n]{\left(x+1\right)\left(x+2\right)\mathrm{\dots }\left(x+n\right)}-x\right)\\ =\underset{x\to 0}{\lim }\left(\sqrt[n]{\left(\frac{1}{y}+1\right)\left(\frac{1}{y}+2\right)\mathrm{\dots }\left(\frac{1}{y}+n\right)}-\frac{1}{y}\right)\\ =\underset{x\to 0}{\lim }\frac{\sqrt[n]{(1+y)(1+2y)\mathrm{\dots }(1+ny)}-1}{y}\end{array}$$\begin{array}{l}*\sqrt[n]{(1+y)(1+2y)\mathrm{\dots }(1+ny)}-1\\ =\sqrt[n]{1+y}-\sqrt[n]{1+y}+\sqrt[n]{(1+y)(1+2y)}-\sqrt[n]{(1+y)(1+2y)+\mathrm{\dots }}\\ -\sqrt[n]{(1+y)(1+2y)\mathrm{\dots }(1+(n-1\left)y\right)}+\sqrt[n]{(1+y)(1+2y)\mathrm{\dots }(1+ny)}-1\\ =\left(\sqrt[n]{1+y}-1\right)+\sqrt[n]{1+y}\left(\sqrt[n]{1+2y}-1\right)+\mathrm{\dots }+\sqrt[n]{(1+y)(1+2y)\mathrm{\dots }(1+(n-1\left)y\right)}\left(\sqrt[n]{1+ny}-1\right)\\ \Rightarrow \underset{y\to 0}{\lim }\frac{\sqrt[n]{(1+y)(1+2y)\mathrm{\dots }(1+ny)}-1}{y}\\ =\underset{y\to 0}{\lim }\left[\frac{\left(\sqrt[n]{1+y}-1\right)}{y}\right]+\underset{y\to 0}{\lim }\left[\sqrt[n]{1+y}\frac{\left(\sqrt[n]{1+2y}-1\right)}{y}\right]+\mathrm{\dots }\\ +\underset{y\to 0}{\lim }\left[\sqrt[n]{(1+y)(1+2y)\mathrm{\dots }(1+(n-1\left)y\right)}.\frac{\left(\sqrt[n]{1+ny}-1\right)}{y}\right]\end{array}$Tổng quát$\begin{array}{l}\underset{y\to 0}{\lim }\left[\sqrt[n]{(1+y)(1+2y)\mathrm{\dots }(1+(k-1)y}.\frac{\sqrt[n]{1+ky}-1}{y}\right]\\ =\underset{y\to 0}{\lim }\left[\sqrt[n]{(1+y)(1+2y)\mathrm{\dots }(1+(k-1)y}.\frac{\left(\sqrt[n]{1+ky}-1\right)\left[{\left(\sqrt[n]{1+ky}\right)}^{n-1}+{\left(\sqrt[n]{1+ky}\right)}^{n-2}+\mathrm{\dots }+1\right]}{y\left[{\left(\sqrt[n]{1+ky}\right)}^{n-1}+{\left(\sqrt[n]{1+ky}\right)}^{n-2}+\mathrm{\dots }+1\right]}\right]\\ =\underset{y\to 0}{\lim }\frac{(1+ky-1).\sqrt[n]{(1+y)(1+2y)\mathrm{\dots }(1+(k-1\left)y\right)}}{{\left(\sqrt[n]{1+ky}\right)}^{n-1}+{\left(\sqrt[n]{1+ky}\right)}^{n-2}+\mathrm{\dots }+1}=\frac{k}{n}\end{array}$Khi đó$\begin{array}{l}\underset{y\to 0}{\lim }\frac{\sqrt[n]{(1+y)(1+2y)\mathrm{\dots }(1+ny)}-1}{y}=\frac{1}{n}+\frac{2}{n}+\frac{3}{n}+\mathrm{\dots }+\frac{n}{n}\\ =\frac{1+2+3+\mathrm{\dots }+n}{n}=\frac{\frac{n(n+1)}{2}}{n}=\frac{n+1}{2}\end{array}$Đáp án cần chọn là: B

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1. Biết rằng a+b=4;limx→1a1−x−b1−x3 hữu hạn. Tính giới hạn L=limx→1b1−x3−a1−x
   
   

   Câu hỏi:
   

   Biết rằng $a+b=4;\underset{x\to 1}{\lim }\left(\frac{a}{1-x}-\frac{b}{1-x^3}\right)$ hữu hạn. Tính giới hạn $L=\underset{x\to 1}{\lim }\left(\frac{b}{1-x^3}-\frac{a}{1-x}\right)$

   A. -1

   B. 2

   C. 1

   Đáp án chính xác

   D. -2

   Trả lời:

   

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2. Cho hàm số f(x)=x2+2x+4−x2−2x+4. Khẳng định nào sau đây là đúng?
   
   

   Câu hỏi:
   

   Cho hàm số $f\left(x\right)=\sqrt{x^2+2x+4}-\sqrt{x^2-2x+4}$. Khẳng định nào sau đây là đúng?

   A. Giới hạn của f(x) khi x→+∞ là 0

   B. Giới hạn của f(x) khi x→-∞ là 2

   C. Giới hạn của f(x) khi x→+∞ là 2

   D. $\underset{x\to -\infty }{\lim }f\left(x\right)=-\underset{x\to -\infty }{\lim }f\left(x\right)$

   Đáp án chính xác

   Trả lời:

   $f\left(x\right)=\sqrt{x^2+2x+4}-\sqrt{x^2-2x+4}$Ta có$\begin{array}{l}\underset{x\to +\infty }{\lim }f\left(x\right)=\underset{x\to +\infty }{\lim }(\sqrt{x^2+2x+4}-\sqrt{x^2-2x+4})\\ =\underset{x\to +\infty }{\lim }\frac{\left(\sqrt{x^2+2x+4}-\sqrt{x^2-2x+4}\right)\left(\sqrt{x^2+2x+4}+\sqrt{x^2-2x+4}\right)}{\left(\sqrt{x^2+2x+4}+\sqrt{x^2-2x+4}\right)}\\ =\underset{x\to +\infty }{\lim }\frac{\left(x^2+2x+4\right)-\left(x^2-2x+4\right)}{\sqrt{x^2+2x+4}+\sqrt{x^2-2x+4}}\\ =\underset{x\to +\infty }{\lim }\frac{4x}{x\sqrt{1+\frac{2}{x}+\frac{4}{x^2}}+x\sqrt{1-\frac{2}{x}+\frac{4}{x^2}}}\\ =\underset{x\to +\infty }{\lim }\frac{4}{\sqrt{1+\frac{2}{x}+\frac{4}{x^2}}+\sqrt{1-\frac{2}{x}+\frac{4}{x^2}}}=2\end{array}$$\begin{array}{l}\underset{x\to -\infty }{\lim }f\left(x\right)=\underset{x\to -\infty }{\lim }\left(\sqrt{x^2+2x+4}-\sqrt{x^2-2x+4}\right)\\ =\underset{x\to -\infty }{\lim }\frac{\left(\sqrt{x^2+2x+4}-\sqrt{x^2-2x+4}\right)\left(\sqrt{x^2+2x+4}+\sqrt{x^2-2x+4}\right)}{\left(\sqrt{x^2+2x+4}+\sqrt{x^2-2x+4}\right)}\\ =\underset{x\to -\infty }{\lim }\frac{\left(x^2+2x+4\right)-\left(x^2-2x+4\right)}{\sqrt{x^2+2x+4}+\sqrt{x^2-2x+4}}\\ =\underset{x\to -\infty }{\lim }\frac{4x}{\sqrt{x^2+2x+4}+\sqrt{x^2-2x+4}}\\ =\underset{x\to -\infty }{\lim }\frac{4x}{-x\sqrt{1+\frac{2}{x}+\frac{4}{x^2}}-x\sqrt{1-\frac{2}{x}+\frac{4}{x^2}}}\\ =\underset{x\to -\infty }{\lim }\frac{4}{-\sqrt{1+\frac{2}{x}+\frac{4}{x^2}}-\sqrt{1-\frac{2}{x}+\frac{4}{x^2}}}=\frac{4}{-1-1}=-2\end{array}$Đáp án cần chọn là: D

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3. Giá trị của giới hạn limx→+∞x2+x−x3−x23
   
   

   Câu hỏi:
   

   Giá trị của giới hạn $\underset{x\to +\infty }{\lim }\left(\sqrt{x^2+x}-\sqrt[3]{x^3-x^2}\right)$

   A. $\frac{5}{6}$

   Đáp án chính xác

   B. $+\infty$

   C. -1

   D. $–\infty$ 

   Trả lời:

   $\begin{array}{l}\underset{x\to +\infty }{\lim }\left(\sqrt{x^2+x}-\sqrt[3]{x^3-x^2}\right)\\ =\underset{x\to +\infty }{\lim }\left(\sqrt{x^2+x}-x+x-\sqrt[3]{x^3-x^2}\right)\\ =\underset{x\to +\infty }{\lim }\left(\frac{x}{\sqrt{x^2+x}+x}+\frac{x^2}{x^2+x\sqrt[3]{x^3-x^2}+\sqrt[3]{{\left(x^3-x^2\right)}^2}}\right)\\ =\underset{x\to +\infty }{\lim }\left(\frac{1}{\sqrt{1+\text{}\frac{1}{x}}+1\text{\hspace{0.17em}\hspace{0.17em}}}+\text{\hspace{0.17em}\hspace{0.17em}}\frac{1}{1+\text{}\sqrt[3]{1-\frac{1}{x}}+\text{}\sqrt[3]{{(1-\frac{1}{x})}^2}\text{\hspace{0.17em}\hspace{0.17em}}}\right)\\ =\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\end{array}$Đáp án cần chọn là: A

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4. Tính limx→−∞x3x+22×3+x2−1
   
   

   Câu hỏi:
   

   Tính $\underset{x\to -\infty }{\lim }x\sqrt{\frac{3x+2}{2x^3+x^2-1}}$

   A. $-\sqrt{\frac{3}{2}}$

   Đáp án chính xác

   B. $\sqrt{\frac{3}{2}}$

   C. $\frac{3}{2}$

   D. $–\frac{3}{2}$ 

   Trả lời:

   $\begin{array}{l}\underset{x\to -\infty }{\lim }x\sqrt{\frac{3x+2}{2x^3+x^2-1}}\\ =\underset{x\to -\infty }{\lim }\left(-\sqrt{\frac{x^2(3x+2)}{2x^3+x^2-1}}\right)\\ =\underset{x\to -\infty }{\lim }\left(-\sqrt{\frac{3x^3+2x^2)}{2x^3+x^2-1}}\right)\\ =\underset{x\to -\infty }{\lim }\left(-\sqrt{\frac{3+\frac{2}{x}}{2+\frac{1}{x}-\frac{1}{x^3}}}\right)=-\sqrt{\frac{3}{2}}\end{array}$Đáp án cần chọn là: A

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5. Tính limx→01+2x.1+3×3.1+4×4−1x
   
   

   Câu hỏi:
   

   Tính $\underset{x\to 0}{\lim }\frac{\sqrt{1+2x}.\sqrt[3]{1+3x}.\sqrt[4]{1+4x}-1}{x}$

   A. $\frac{23}{2}$

   B. 24

   C. $\frac{3}{2}$ 

   D. 3

   Đáp án chính xác

   Trả lời:

   $\begin{array}{l}\sqrt{1+2x}.\sqrt[3]{1+3x}.\sqrt[4]{1+4x}-1\\ =\sqrt{1+2x}-\sqrt{1+2x}+\sqrt{1+2x}.\sqrt[3]{1+3x}-\sqrt{1+2x}.\sqrt[3]{1+3x}+\sqrt{1+2x}.\sqrt[3]{1+3x}.\sqrt[4]{1+4x}-1\\ =\left(\sqrt{1+2x}-1\right)+\sqrt{1+2x}\left(\sqrt[3]{1+3x}-1\right)+\sqrt{1+2x}.\sqrt[3]{1+3x}.\left(\sqrt[4]{1+4x}-1\right)\\ \Rightarrow \underset{x\to 0}{\lim }\frac{\sqrt{1+2x}.\sqrt[3]{1+3x}.\sqrt[4]{1+4x}-1}{x}\\ =\underset{x\to 0}{\lim }\left(\frac{\sqrt{1+2x}-1}{x}\right)+\underset{x\to 0}{\lim }\left(\sqrt{1+2x}.\frac{\sqrt[3]{1+3x}-1}{x}\right)+\underset{x\to 0}{\lim }\left(\sqrt{1+2x}.\sqrt[3]{1+3x}.\frac{\sqrt[4]{1+4x}-1}{x}\right)\end{array}$Tính $\begin{array}{l}\underset{x\to 0}{\lim }\left(\frac{\sqrt{1+2x}-1}{x}\right)=\underset{x\to 0}{\lim }\left(\frac{\left(\sqrt{1+2x}-1\right)\left(\sqrt{1+2x}+1\right)}{x\left(\sqrt{1+2x}+1\right)}\right)\\ =\underset{x\to 0}{\lim }\frac{2x}{x\left(\sqrt{1+2x}+1\right)}=\underset{x\to 0}{\lim }\frac{2}{\sqrt{1+2x}+1}=\frac{2}{1+1}=1\end{array}$$\begin{array}{l}\underset{x\to 0}{\lim }\left(\sqrt{1+2x.}\frac{\left(\sqrt[3]{1+3x}-1\right)}{x}\right)\\ =\underset{x\to 0}{\lim }\left(\sqrt{1+2x.}\frac{\left(\sqrt[3]{1+3x}-1\right)\left[{\left(\sqrt[3]{1+3x}\right)}^2+\sqrt[3]{1+3x}+1\right]}{x\left[{\left(\sqrt[3]{1+3x}\right)}^2+\sqrt[3]{1+3x}+1\right]}\right)\\ =\underset{x\to 0}{\lim }\left(\sqrt{1+2x.}\frac{3x}{x\left[{\left(\sqrt[3]{1+3x}\right)}^2+\sqrt[3]{1+3x}+1\right]}\right)\\ =\underset{x\to 0}{\lim }\left(\frac{3\sqrt{1+2x}}{\left[{\left(\sqrt[3]{1+3x}\right)}^2+\sqrt[3]{1+3x}+1\right]}\right)=\frac{3.1}{1+1+1}=1\end{array}$$\begin{array}{l}\underset{x\to 0}{\lim }\left(\sqrt{1+2x}.\sqrt[3]{1+3x}.\frac{\left(\sqrt[4]{1+4x}-1\right)}{x}\right)\\ =\underset{x\to 0}{\lim }\left(\sqrt{1+2x}.\sqrt[3]{1+3x}.\frac{\left(\sqrt[4]{1+4x}-1\right)\left[{\left(\sqrt[4]{1+4x}\right)}^3+{\left(\sqrt[4]{1+4x}\right)}^2+\sqrt[4]{1+4x}+1\right]}{x\left[{\left(\sqrt[4]{1+4x}\right)}^3+{\left(\sqrt[4]{1+4x}\right)}^2+\sqrt[4]{1+4x}+1\right]}\right)\\ =\underset{x\to 0}{\lim }\left(\sqrt{1+2x}.\sqrt[3]{1+3x}.\frac{4x}{x\left[{\left(\sqrt[4]{1+4x}\right)}^3+{\left(\sqrt[4]{1+4x}\right)}^2+\sqrt[4]{1+4x}+1\right]}\right)\\ =\underset{x\to 0}{\lim }\frac{4\sqrt{1+2x}.\sqrt[3]{1+3x}.}{{\left(\sqrt[4]{1+4x}\right)}^3+{\left(\sqrt[4]{1+4x}\right)}^2+\sqrt[4]{1+4x}+1}\\ =\frac{\mathrm{4.1.1}}{1+1+1+1}=1\end{array}$Vậy $\underset{x\to 0}{\lim }\frac{\sqrt{1+2x}.\sqrt[3]{1+3x}.\sqrt[4]{1+4x}-1}{x}=1+1+1=3$Đáp án cần chọn là: D

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