Đáp án:
\(\begin{array}{l}
5)\quad\lim\limits_{(x,y)\to (+\infty,+\infty)}(x^2 + y^2)e^{-x^2 - y^2} =0\\
7)\quad\lim\limits_{(x,y)\to (0;0)}\dfrac{xy^2}{2 - \sqrt{4 + xy^2}} = -4
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
5)\quad \lim\limits_{(x,y)\to (+\infty,+\infty)}(x^2 + y^2)e^{-x^2 - y^2}\\
\text{Chọn}\ \{M_n(n;n)\}.\ \text{Ta có:}\ \lim\limits_{n\to +\infty}M_n= (+\infty,+\infty)\\
\quad \lim\limits_{n\to +\infty}(n^2 + n^2)e^{-n^2 - n^2}\\
= \lim\limits_{n\to +\infty}2n^2e^{-2n^2}\\
= 2.0 = 0\\
\text{Chọn}\ \{N_n(n;2n)\}.\ \text{Ta có:}\ \lim\limits_{n\to +\infty}N_n= (+\infty,+\infty)\\
\quad \lim\limits_{n\to +\infty}(n^2 + 4n^2)e^{-n^2 - 4n^2}\\
= \lim\limits_{n\to +\infty}5n^2e^{-5n^2}\\
= 5.0 = 0\\
\text{Vậy}\ \lim\limits_{(x,y)\to (+\infty,+\infty)}(x^2 + y^2)e^{-x^2 - y^2} =0\\
7)\quad \lim\limits_{(x,y)\to (0;0)}\dfrac{xy^2}{2 - \sqrt{4 + xy^2}}\\
= -\lim\limits_{(x,y)\to (0;0)}\left(2 + \sqrt{4 + xy^2}\right)\\
\text{Chọn}\ \left\{M_n\left(\dfrac1n;\dfrac{1}{\sqrt n}\right)\right\}.\ \text{Ta có:}\ \lim\limits_{n\to +\infty}M_n= (0,0)\\
\quad -\lim\limits_{n\to +\infty}\left(2 + \sqrt{4 + \dfrac1n\cdot \dfrac{1}{n}}\right)\\
= - \left(2 + \sqrt{4 + 0}\right)\\
= - 4\\
\text{Chọn}\ \left\{N_n\left(\dfrac1n;\dfrac{2}{\sqrt n}\right)\right\}.\ \text{Ta có:}\ \lim\limits_{n\to +\infty}N_n= (0,0)\\
\quad -\lim\limits_{n\to +\infty}\left(2 + \sqrt{4 + \dfrac1n\cdot \dfrac{4}{n}}\right)\\
= - \left(2 + \sqrt{4 + 0}\right)\\
= - 4\\
\text{Vậy}\ \lim\limits_{(x,y)\to (0;0)}\dfrac{xy^2}{2 - \sqrt{4 + xy^2}} = -4
\end{array}\)
