Đáp án:
\(\begin{array}{l}
1)\quad \lim\dfrac{3^n -4^{n+1}}{2^{2n} + 10.3^n + 7}=-4\\
2)\quad \lim\limits_{x\to 1}\dfrac{x^2 - 3x + 2}{x-1}=-1\\
3)\quad \lim\limits_{x\to +\infty}\dfrac{2x^3 + 3 x- 4}{-x^3 - x^2 + 1}=-2\\
4)\quad \lim\limits_{x\to 3}\dfrac{\sqrt{x+1} - 2}{x^2 - 9}=\dfrac{1}{24}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)\quad \lim\dfrac{3^n -4^{n+1}}{2^{2n} + 10.3^n + 7}\\
= \lim\dfrac{3^n - 4.4^n}{4^n + 10.3^n + 7}\\
= \lim\dfrac{\left(\dfrac34\right)^n - 4}{1 + 10.\left(\dfrac34\right)^n + 7.\left(\dfrac14\right)^n}\\
= \dfrac{0 -4}{1 +10.0 + 7.0}\\
= -4\\
2)\quad \lim\limits_{x\to 1}\dfrac{x^2 - 3x + 2}{x-1}\\
= \lim\limits_{x\to 1}\dfrac{(x-1)(x-2)}{x-1}\\
= \lim\limits_{x\to 1}(x-2)\\
= 1 - 2\\
= -1\\
3)\quad \lim\limits_{x\to +\infty}\dfrac{2x^3 + 3 x- 4}{-x^3 - x^2 + 1}\\
=\lim\limits_{x\to +\infty}\dfrac{2 + \dfrac{3}{x^2} - \dfrac{4}{x^3}}{-1 - \dfrac1x + \dfrac{1}{x^3}}\\
= \dfrac{2 +0 -0}{-1 - 0 + 0}\\
= -2\\
4)\quad \lim\limits_{x\to 3}\dfrac{\sqrt{x+1} - 2}{x^2 - 9}\\
= \lim\limits_{x\to 3}\dfrac{(\sqrt{x+1} -2)(\sqrt{x+1} + 2)}{(x-3)(x+3)(\sqrt{x+1} +2)}\\
= \lim\limits_{x\to 3}\dfrac{1}{(x+3)(\sqrt{x+1} +2)}\\
= \dfrac{1}{(3+3)(\sqrt{3+1} +2)}\\
=\dfrac{1}{24}
\end{array}\)
