c) $C=\lim _{x \rightarrow(-2)^{-}} \frac{\sqrt{x^{2}-4}}{x^{2}+2 x}$.

Q: c) $C=\lim _{x \rightarrow(-2)^{-}} \frac{\sqrt{x^{2}-4}}{x^{2}+2 x}$.

<p><span class="math-tex">$\begin{array}{l}\text {} C=\lim _{x \rightarrow-2^{-}} \frac{\sqrt{x^{2}-4}}{x^{2}+2 x} . \\ =\lim _{x \rightarrow(-2)^{-}} \frac{\sqrt{2-x} \cdot \sqrt{-2-x}}{-x(-2-x)}=\lim _{x \rightarrow(-2)^{-}} \frac{\sqrt{2-x}}{-x \sqrt{-2-x}}=+\infty\end{array}$</span></p>

Question

Accepted Answer

$\begin{array}{l}\text {} C=\lim _{x \rightarrow-2^{-}} \frac{\sqrt{x^{2}-4}}{x^{2}+2 x} . \ =\lim _{x \rightarrow(-2)^{-}} \frac{\sqrt{2-x} \cdot \sqrt{-2-x}}{-x(-2-x)}=\lim _{x \rightarrow(-2)^{-}} \frac{\sqrt{2-x}}{-x \sqrt{-2-x}}=+\infty\end{array}$

THPT Nguyễn Thị Minh Khai - Đề thi cuối kì 2 (CT) 18-19 - Q. 3 - Tp. Hồ Chí Minh - MĐ 6633