Giải Toán 11 SGK nâng cao Chương 4 Luyện tập (trang 167)

Tìm các giới hạn sau:

a) $\underset{x\to 0}{lim}\left(\frac{1}{x}+\frac{1}{x^2}\right)$

b) $\underset{x\to -2}{lim}\frac{x^3+8}{x+2}$

c) $\underset{x\to 9}{lim}\frac{3-\sqrt{x}}{9-x}$

d) $\underset{x\to 0}{lim}\frac{2-\sqrt{4-x}}{x}$

e) $\underset{x\to +\mathrm{\infty }}{lim}\frac{x^4-x^3+11}{2x-7}$

f) $\underset{x\to -\mathrm{\infty }}{lim}\frac{\sqrt{x^4+4}}{x+4}$

Hướng dẫn giải:

Câu a:

$\underset{x\to 0}{lim}\left(\frac{1}{x}+\frac{1}{x^2}\right)=\underset{x\to 0}{lim}\frac{x+1}{x^2}=+\mathrm{\infty }$ (vì $\underset{x\to 0}{lim}\left(x+1\right)=1>0,\underset{x\to 0}{lim}{x}^2=0$ và $x^2>0,\mathrm{\forall }x\ne 0$)

Câu b:

$\underset{x\to -2}{lim}\frac{x^3+8}{x+2}=\underset{x\to -2}{lim}\frac{\left(x+2\right)\left(x^2-2x+4\right)}{x+2}=\underset{x\to -2}{lim}\left(x^2-2x+4\right)=12$

Câu c:

$\underset{x\to 9}{lim}\frac{3-\sqrt{x}}{9-x}=\underset{x\to 9}{lim}\frac{1}{3+\sqrt{x}}=\frac{1}{6}$

Câu d:

$\underset{x\to 0}{lim}\frac{2-\sqrt{4-x}}{x}=\underset{x\to 0}{lim}\frac{4-\left(4-x\right)}{x\left(2+\sqrt{4-x}\right)}=\underset{x\to 0}{lim}\frac{1}{2+\sqrt{4-x}}=\frac{1}{4}$

Câu e:

$\underset{x\to +\mathrm{\infty }}{lim}\frac{x^4-x^3+11}{2x-7}=\underset{x\to +\mathrm{\infty }}{lim}\frac{x^3-x^2+\frac{11}{x}}{2-\frac{7}{x}}=+\mathrm{\infty }$

Câu f:

Với x < 0, ta có: $\frac{\sqrt{x^4+4}}{x+4}=\frac{x^2\sqrt{1+\frac{4}{x^4}}}{x+4}=\frac{x\sqrt{1+\frac{4}{x^4}}}{1+\frac{4}{x}}$

Vì $\underset{x\to -\mathrm{\infty }}{lim}x\sqrt{1+\frac{4}{x^4}}=-\mathrm{\infty },\underset{x\to -\mathrm{\infty }}{lim}\left(1+\frac{4}{x}\right)=1>0$ nên $\underset{x\to -\mathrm{\infty }}{lim}\frac{\sqrt{x^4+4}}{x+4}=-\mathrm{\infty }$

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Tìm các giới hạn sau:

a) $\underset{x\to -\sqrt{3}}{lim}\frac{x^3+3\sqrt{3}}{3-x^2}$

b) $\underset{x\to 4}{lim}\frac{\sqrt{x}-2}{x^2-4x}$

c) $\underset{x\to 1^{+}}{lim}\frac{\sqrt{x-1}}{x^2-x}$

d) $\underset{x\to 0}{lim}\frac{\sqrt{x^2+x+1}-1}{3x}$

Hướng dẫn giải:

Câu a:

Ta có $\frac{x^3+3\sqrt{3}}{3-x^2}=\frac{\left(x+\sqrt{3}\right)\left(x^2-x\sqrt{3}+3\right)}{\left(x+\sqrt{3}\right)\left(\sqrt{3}-x\right)}=\frac{x^2-x\sqrt{3}+3}{\sqrt{3}-x}$

Do đó $\underset{x\to -\sqrt{3}}{lim}\frac{x^3+3\sqrt{3}}{3-x^2}=\underset{x\to -\sqrt{3}}{lim}\frac{x^2-x\sqrt{3}+3}{\sqrt{3}-x}=\frac{9}{2\sqrt{3}}=\frac{3\sqrt{3}}{2}$

Câu b:

$\underset{x\to 4}{lim}\frac{\sqrt{x}-2}{x^2-4x}=\underset{x\to 4}{lim}\frac{\sqrt{x}-2}{x\left(x-4\right)}=\underset{x\to 4}{lim}\frac{1}{x\left(\sqrt{x}+2\right)}=\frac{1}{16}$

Câu c:

$\underset{x\to 1^{+}}{lim}\frac{\sqrt{x-1}}{x^2-x}=\underset{x\to 1^{+}}{lim}\frac{\sqrt{x-1}}{x\left(x-1\right)}=\underset{x\to 1^{+}}{lim}\frac{1}{x\sqrt{x-1}}=+\mathrm{\infty }$

Câu d:

$\underset{x\to 0}{lim}\frac{\sqrt{x^2+x+1}-1}{3x}=\underset{x\to 0}{lim}\frac{x^2+x+1-1}{3x\left(\sqrt{x^2+x+1}+1\right)}=\frac{1}{3}\underset{x\to 0}{lim}\frac{x+1}{\sqrt{x^2+x+1}+1}=\frac{1}{6}$

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Tìm các giới hạn sau:

a) $\underset{x\to -\mathrm{\infty }}{lim}x\sqrt{\frac{2x^3+x}{x^5-x^2+3}}$

b) $\underset{x\to -\mathrm{\infty }}{lim}\frac{\left|x\right|+\sqrt{x^2+x}}{x+10}$

c) $\underset{x\to +\mathrm{\infty }}{lim}\frac{\sqrt{2x^4+x^2-1}}{1-2x}$

d) $\underset{x\to -\mathrm{\infty }}{lim}\left(\sqrt{2x^2+1}+x\right)$

Hướng dẫn giải:

Câu a:

Với x < 0, ta có:

$\begin{array}{l}x\sqrt{\frac{2x^3+x}{x^5-x^2+3}}=-\left|x\right|\sqrt{\frac{2x^3+x}{x^5-x^2+3}}\\ =-\sqrt{\frac{x^2\left(2x^3+x\right)}{x^5-x^2+3}}=-\sqrt{\frac{2+\frac{1}{x^2}}{1-\frac{1}{x^3}+\frac{1}{x^5}}}\end{array}$

Do đó $\underset{x\to -\mathrm{\infty }}{lim}x\sqrt{\frac{2x^3+x}{x^5-x^2+3}}=-\sqrt{2}$

Câu b:

$\begin{array}{l}\underset{x\to -\mathrm{\infty }}{lim}\frac{\left|x\right|+\sqrt{x^2+x}}{x+10}=\underset{x\to -\mathrm{\infty }}{lim}\frac{\left|x\right|+\left|x\right|\sqrt{1+\frac{1}{x}}}{x+10}\\ =\underset{x\to -\mathrm{\infty }}{lim}\frac{-x-x\sqrt{1+\frac{1}{x}}}{x+10}=\underset{x\to -\mathrm{\infty }}{lim}\frac{-1-\sqrt{1+\frac{1}{x}}}{1+\frac{10}{x}}=-2\end{array}$

Câu c:

$\underset{x\to +\mathrm{\infty }}{lim}\frac{\sqrt{2x^4+x^2-1}}{1-2x}=\underset{x\to +\mathrm{\infty }}{lim}\frac{x^2\sqrt{2+\frac{1}{x^2}-\frac{1}{x^4}}}{x\left(\frac{1}{x}-2\right)}=\underset{x\to +\mathrm{\infty }}{lim}x.\frac{\sqrt{2+\frac{1}{x^2}-\frac{1}{x^4}}}{\frac{1}{x}-2}=-\mathrm{\infty }$

(vì $\underset{x\to +\mathrm{\infty }}{lim}x=+\mathrm{\infty },\underset{x\to +\mathrm{\infty }}{lim}\frac{\sqrt{2+\frac{1}{x^2}-\frac{1}{x^4}}}{\frac{1}{x}-2}=-\frac{\sqrt{2}}{2}<0$)

Câu d:

$\begin{array}{l}\underset{x\to -\mathrm{\infty }}{lim}\left(\sqrt{2x^2+1}+x\right)=\underset{x\to -\mathrm{\infty }}{lim}\frac{2x^2+x-x^2}{\sqrt{2x^2+x}-x}\\ =\underset{x\to -\mathrm{\infty }}{lim}\frac{x\left(x+1\right)}{-x\left(\sqrt{2+\frac{1}{x}}+1\right)}=\underset{x\to -\mathrm{\infty }}{lim}-\frac{x+1}{\sqrt{2+\frac{1}{x}+1}}=+\mathrm{\infty }\end{array}$

(vì $\underset{x\to -\mathrm{\infty }}{lim}\left(-x-1\right)=+\mathrm{\infty }$)

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Tìm các giới hạn sau:

a) $\underset{x\to 0^{+}}{lim}\frac{\sqrt{x^2+x}-\sqrt{x}}{x^2}$

b) $\underset{x\to 1^{-}}{lim}x.\frac{\sqrt{1-x}}{2\sqrt{1-x}+1-x}$

c) $\underset{x\to 3^{-}}{lim}\frac{3-x}{\sqrt{27-x^3}}$

d) $\underset{x\to 2^{+}}{lim}\frac{\sqrt{x^3-8}}{x^2-2x}$

Hướng dẫn giải:

Câu a:

$\underset{x\to 0^{+}}{lim}\frac{\sqrt{x^2+x}-\sqrt{x}}{x^2}=\underset{x\to 0^{+}}{lim}\frac{x^2}{x^2\left(\sqrt{x^2+x}+\sqrt{x}\right)}=\underset{x\to 0^{+}}{lim}\frac{1}{\sqrt{x^2+x}+\sqrt{x}}=+\mathrm{\infty }$

Câu b:

$\underset{x\to 1^{-}}{lim}x.\frac{\sqrt{1-x}}{2\sqrt{1-x}+1-x}=\underset{x\to 1^{-}}{lim}\frac{x}{2+\sqrt{1-x}}=\frac{1}{2}$

Câu c:

$\begin{array}{l}\underset{x\to 3^{-}}{lim}\frac{3-x}{\sqrt{27-x^3}}=\underset{x\to 3^{-}}{lim}\frac{{\left(\sqrt{3-x}\right)}^2}{\sqrt{\left(3-x\right)\left(x^2+3x+9\right)}}\\ =\underset{x\to 3^{-}}{lim}\frac{\sqrt{3-x}}{\sqrt{x^2+3x+9}}=0\end{array}$

Câu d:

$\underset{x\to 2^{+}}{lim}\frac{\sqrt{x^3-8}}{x^2-2x}=\underset{x\to 2^{+}}{lim}\frac{\sqrt{\left(x-2\right)\left(x^2+2x+4\right)}}{x\left(x-2\right)}=\underset{x\to 2^{+}}{lim}\frac{1}{x}\sqrt{\frac{x^2+2x+4}{x-2}}=+\mathrm{\infty }$

(vì $\underset{x\to 2^{+}}{lim}\sqrt{x^2+2x+4}=2\sqrt{3},\underset{x\to 2^{+}}{lim}x\sqrt{x-2}=0$ và $x\sqrt{x-2}>0,\mathrm{\forall }x>2$).

 

Trên đây là nội dung chi tiết Giải bài tập nâng cao Toán 11 Chương 4 Luyện tập (trang 167) với hướng dẫn giải chi tiết, rõ ràng, trình bày khoa học. Chúng tôi hy vọng đây sẽ là tài liệu hữu ích giúp các bạn học sinh lớp 11 học tập thật tốt.