41 câu Dạng 2 : Tìm giới hạn của hàm số dạng vô định có đáp án

Hướng dẫn giải

Ta có $\underset{x\to 5}{\lim }\frac{\sqrt{3x+1}-4}{3-\sqrt{x+4}}=\underset{x\to 5}{\lim }\frac{\left[\left(3x+1\right)-16\right]\left(3+\sqrt{x+4}\right)}{\left[9-\left(x+4\right)\right]\left(\sqrt{3x+1}+4\right)}$.

$=\underset{x\to 5}{\lim }\frac{-3\left(3+\sqrt{x+4}\right)}{\sqrt{3x+1}+4}=\frac{-18}{8}=-\frac{9}{4}$

Hướng dẫn giải

Ta có $M=\underset{x\to 0}{\lim }\frac{\sqrt{4x+1}-\left(2x+1\right)}{x^2}-\underset{x\to 0}{\lim }\frac{\sqrt[3]{1+6x}-\left(2x+1\right)}{x^2}$

$=\underset{x\to 0}{\lim }\frac{-4}{\sqrt{4x+1}+2x+1}-\underset{x\to 0}{\lim }\frac{-8x-12}{\sqrt[3]{{\left(1+6x\right)}^2}+\left(2x+1\right)\sqrt[3]{1+6x}+{\left(2x+1\right)}^2}$

$=-2+4=2$

Hướng dẫn giải

Ta có $\sqrt{1+ax}\sqrt[3]{1+bx}\sqrt[4]{1+cx}-1$.

$=\sqrt{1+ax}\sqrt[3]{1+bx}\left(\sqrt[4]{1+cx}-1\right)+\sqrt{1+ax}\left(\sqrt[3]{1+bx}-1\right)+\left(\sqrt{1+ax}-1\right)$

$B=\underset{x\to 0}{\lim }\left(\sqrt{1+ax}\sqrt[3]{1+bx}\right)\frac{\sqrt[4]{1+cx}-1}{x}+\underset{x\to 0}{\lim }\sqrt{1+ax}\frac{\sqrt[3]{1+bx}-1}{x}+\underset{x\to 0}{\lim }\frac{\sqrt{1+ax}-1}{x}$

$\Rightarrow B=\frac{c}{4}+\frac{b}{3}+\frac{a}{2}$

Hướng dẫn giải

Theo giả thiết có $\underset{x\to 2}{\lim }\left(f\left(x\right)-16\right)=0\iff \underset{x\to 2}{\lim }f\left(x\right)=16$ .

Ta có $\underset{x\to 2}{\lim }\frac{\sqrt[3]{5f\left(x\right)-16}-4}{x^2+2x-8}$

$=\underset{x\to 2}{\lim }\frac{\left(5f\left(x\right)-16\right)-64}{\left(x-2\right)\left(x+4\right)\left[{\left(\sqrt[3]{5f\left(x\right)-16}\right)}^2+4\sqrt[3]{5f\left(x\right)-16}+4^2\right]}$

$=\underset{x\to 2}{\lim }\frac{5\left(f\left(x\right)-16\right)}{\left(x-2\right)\left(x+4\right)\left[{\left(\sqrt[3]{5f\left(x\right)-16}\right)}^2+4\sqrt[3]{5f\left(x\right)-16}+4^2\right]}$

$=\underset{x\to 2}{\lim }\left[\frac{f\left(x\right)-16}{x-2}.\frac{5}{\left(x+4\right)\left({\left(\sqrt[3]{5f\left(x\right)-16}\right)}^2+4\sqrt[3]{5f\left(x\right)-16}+4^2\right)}\right]$

$=12.\frac{5}{6\left[{\left(\sqrt[3]{5.16-16}\right)}^2+4\sqrt[3]{5.16-16}+16\right]}=\frac{5}{24}$

Ta có: $\underset{x\to 2}{\lim }\frac{\sqrt[3]{x^2+4}-2}{x^2-4}=\underset{x\to 2}{\lim }\frac{x^2-4}{\left(\sqrt[3]{{\left(x^2+4\right)}^2}+2\sqrt[3]{x^2+4}+4\right)\left(x^2-4\right)}$

$=\underset{x\to 2}{\lim }\frac{1}{\left(\sqrt[3]{{\left(x^2+4\right)}^2}+2\sqrt[3]{x^2+4}+4\right)}=\frac{1}{12}$

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

Ta có: $\underset{x\to 3}{\lim }\frac{x^4-27x}{2x^2-3x-9}=\underset{x\to 3}{\lim }\frac{x\left(x-3\right)\left(x^2+3x+9\right)}{\left(2x+3\right)\left(x-3\right)}=\underset{x\to 3}{\lim }\frac{x\left(x^2+3x+9\right)}{\left(2x+3\right)}=9$

Ta có: $\underset{x\to 1}{\lim }\frac{2x-\sqrt{3x+1}}{x^2-1}=\underset{x\to 1}{\lim }\frac{4x^2-3x-1}{\left(x^2-1\right)\left(2x+\sqrt{3x+1}\right)}=\underset{x\to 1}{\lim }\frac{4x+1}{\left(x+1\right)\left(2x+\sqrt{3x+1}\right)}=\frac{5}{8}$

Ta có: $\underset{x\to -1}{\lim }\frac{\sqrt[3]{x}+1}{\sqrt{x^2+3}-2}=\underset{x\to -1}{\lim }\frac{\left(x+1\right)\left(\sqrt{x^2+3}+2\right)}{\left(\sqrt[3]{x^2}-\sqrt[3]{x}+1\right)\left(x^2-1\right)}=\underset{x\to -1}{\lim }\frac{\left(\sqrt{x^2+3}+2\right)}{\left(\sqrt[3]{x^2}-\sqrt[3]{x}+1\right)\left(x-1\right)}=\frac{4}{-6}=\frac{-2}{3}$

Ta có: $\underset{x\to 0}{\lim }\frac{{\left(x+a\right)}^3-a^3}{x}=\underset{x\to 0}{\lim }\frac{x^3+3x^{2}a+3x{a}^2}{x}=\underset{x\to 0}{\lim }\left(x^2+3xa+3a^2\right)=3a^2$

Ta có: $\underset{x\to -2}{\lim }\frac{x^4-16}{x^2+6x+8}=\underset{x\to -2}{\lim }\frac{\left(x^2+4\right)\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(x+4\right)}=\underset{x\to -2}{\lim }\frac{\left(x^2+4\right)\left(x-2\right)}{\left(x+4\right)}=-16$

Ta có: $\underset{x\to -2}{\lim }\frac{x^4+8x}{x^3+2x^2+x+2}=\underset{x\to -2}{\lim }\frac{x\left(x+2\right)\left(x^2-2x+4\right)}{\left(x+2\right)\left(x^2+1\right)}=\underset{x\to -2}{\lim }\frac{x\left(x^2-2x+4\right)}{\left(x^2+1\right)}=\frac{-24}{5}$

Ta có: $\underset{x\to -1}{\lim }\frac{\sqrt{x^2+8}-3}{\sqrt{1-x}-\sqrt{2}}=\underset{x\to -1}{\lim }\frac{\left(x^2-1\right)\left(\sqrt{1-x}+\sqrt{2}\right)}{\left(\sqrt{x^2+8}+3\right)\left(-x-1\right)}=\underset{x\to -1}{\lim }\frac{\left(x-1\right)\left(\sqrt{1-x}+\sqrt{2}\right)}{-\left(\sqrt{x^2+8}+3\right)}=\frac{2\sqrt{2}}{3}$

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

Ta có: $\underset{x\to 0}{\lim }\frac{\sqrt{x^2+1}-1}{4-\sqrt{x^2+16}}=\underset{x\to 0}{\lim }\frac{x^2\left(4+\sqrt{x^2+16}\right)}{-x^2\left(\sqrt{x^2+1}+1\right)}=\underset{x\to 0}{\lim }\frac{\left(4+\sqrt{x^2+16}\right)}{-\left(\sqrt{x^2+1}+1\right)}=-4$

Ta có: $\underset{x\to 1}{\lim }\frac{x^m-x^n}{x-1}=\underset{x\to 1}{\lim }\frac{\left(x^m-1\right)-\left(x^n-1\right)}{x-1}$

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

$=\underset{x\to 1}{\lim }\left[\left(x^{m-1}+x^{m-2}+\mathrm{...}+1\right)-\left(x^{n-1}+x^{n-2}+\mathrm{...}+1\right)\right]=m-n$

Ta có: $\underset{x\to 1}{\lim }\frac{\sqrt{2x-1}-1}{x-1}=\underset{x\to 1}{\lim }\frac{2}{\sqrt{2x-1}+1}=1$ và $\underset{x\to 1}{\lim }\frac{\sqrt[3]{3x-2}-1}{x-1}=\underset{x\to 1}{\lim }\frac{3}{\left(\sqrt[3]{{\left(3x-2\right)}^2}+\sqrt[3]{3x-2}+1\right)}=1$

Do đó $L=\underset{x\to 1}{\lim }\frac{\sqrt{2x-1}-\sqrt[3]{3x-2}}{x-1}=\underset{x\to 1}{\lim }\frac{\sqrt{2x-1}-1}{x-1}-\underset{x\to 1}{\lim }\frac{\sqrt[3]{3x-2}-1}{x-1}=1-1=0$ Vậy L=0