Đáp án:

Giải thích các bước giải:

 Câu 1:

a)$=\dfrac{\lim _{x\to \infty \:}\left(3+\dfrac{2}{x}+\dfrac{5}{x^2}\right)}{\lim _{x\to \infty \:}\left(7+\dfrac{1}{x}-\dfrac{8}{x^2}\right)}$

`=3/7`

b) $\lim _{x\to \:3}\left(\dfrac{2x^2-7x+3}{x^2-9}\right)$

$=\lim _{x\to \:3}\left(\dfrac{2x-1}{x+3}\right)$

$=\dfrac{2\cdot \:3-1}{3+3}$

`=5/6`

c) $=\dfrac{\sqrt[3]{3^2+4}-2}{3^2-4}$

$=\dfrac{\sqrt[3]{13}-2}{5}$

Câu 2:)$\lim _{x\to 2}\left(3m-2x\right)=3m-4$

$\lim _{x\to 2}\left(\dfrac{2x-4}{\:\sqrt{4x+1}-3}\right)$

$=\lim _{x\to \:2}\dfrac{2x\left(1-\dfrac{4}{2x}\right)}{\sqrt{4x+1}-3}$

$=2\cdot \lim _{x\to \:2}\left(\dfrac{x-2}{\sqrt{4x+1}-3}\right)$

$=2\cdot \lim _{x\to \:2}\left(\dfrac{\sqrt{4x+1}+3}{4}\right)$

$=2\cdot \dfrac{\sqrt{4\cdot \:2+1}+3}{4}$

$=3$

`=> 3m-4=3 =>m=7/3`

Đáp án:

\(\begin{array}{l}
1)\\
a)\quad \lim\dfrac{3n^2 + 2n+5}{7n^2 +n -8}= \dfrac37\\
b)\quad \lim\limits_{x \to 3}\dfrac{2x^2 - 7x + 3}{x^2 - 9}= \dfrac56\\
c)\quad \lim\limits_{x\to 2}\dfrac{\sqrt[3]{x^2 + 4} - 2}{x^2 - 4}= \dfrac{1}{12}\\
2)\quad m = \dfrac73
\end{array}\) 

Giải thích các bước giải:

 \(\begin{array}{l}
1)\\
a)\quad \lim\dfrac{3n^2 + 2n+5}{7n^2 +n -8}\\
= \lim\dfrac{3 + \dfrac2n + \dfrac{5}{n^2}}{7 + \dfrac1n - \dfrac{8}{n^2}}\\
= \dfrac{3 + 0 +0}{7 + 0 - 0}\\
= \dfrac37\\
b)\quad \lim\limits_{x \to 3}\dfrac{2x^2 - 7x + 3}{x^2 - 9}\\
= \lim\limits_{x \to 3}\dfrac{(2x-1)(x-3)}{(x-3)(x+3)}\\
= \lim\limits_{x \to 3}\dfrac{2x-1}{x+3}\\
= \dfrac{2.3 - 1}{3 + 3}\\
= \dfrac56\\
c)\quad \lim\limits_{x\to 2}\dfrac{\sqrt[3]{x^2 + 4} - 2}{x^2 - 4}\\
= \lim\limits_{x\to 2}\dfrac{\left(\sqrt[3]{x^2 + 4} - 2\right)\left(\sqrt[3]{(x^2 + 4)^2} + 2\sqrt[3]{x^2 + 4} + 4\right)}{(x^2 -4)\left(\sqrt[3]{(x^2 + 4)^2} + 2\sqrt[3]{x^2 + 4} + 4\right)}\\
= \lim\limits_{x\to 2}\dfrac{x^2 + 4 - 8}{(x^2 -4)\left(\sqrt[3]{(x^2 + 4)^2} + 2\sqrt[3]{x^2 + 4} + 4\right)}\\
= \lim\limits_{x\to 2}\dfrac{1}{\sqrt[3]{(x^2 + 4)^2} + 2\sqrt[3]{x^2 + 4} + 4}\\
= \dfrac{1}{\sqrt[3]{(2^2 + 4)^2} + 2\sqrt[3]{2^2 + 4} + 4}\\
= \dfrac{1}{12}\\
2)\\
\quad f(x)= \begin{cases}3m - 2x\quad \qquad khi\quad x =2\\\dfrac{2x-4}{\sqrt{4x+1} -3}\quad khi \quad x \ne 3\end{cases}\\
\text{Ta có:}\\
+)\quad f(2) = 3m -2.2 = 3m - 4\\
+)\quad \lim\limits_{x\to 2}f(x)\\
= \lim\limits_{x\to 2}\dfrac{2x-4}{\sqrt{4x+1} -3}\\
= \lim\limits_{x\to 2}\dfrac{(2x-4)\left(\sqrt{4x+1} +3\right)}{\left(\sqrt{4x+1} +3\right)\left(\sqrt{4x+1} -3\right)}\\
= \lim\limits_{x\to 2}\dfrac{(2x-4)\left(\sqrt{4x+1} +3\right)}{4x + 1 -9}\\
= \lim\limits_{x\to 2}\dfrac{\sqrt{4x+1} +3}{2}\\
= \dfrac{\sqrt{4.2 +1} +3}{2}\\
= 3\\
\text{Hàm số liên tục tại $x_o= 2$}\\
\Leftrightarrow f(2) = \lim\limits_{x\to 2}f(x)\\
\Leftrightarrow 3m -4 = 3\\
\Leftrightarrow m = \dfrac73\\
\text{Vậy}\ m = \dfrac73
\end{array}\)