Đáp án:
\(\lim\limits_{x\to -2}\dfrac{\sqrt{x+6} +x}{x^2-4}= -\dfrac{5}{16}\)
Giải thích các bước giải:
\(\begin{array}{l}
\quad \lim\limits_{x\to -2}\dfrac{\sqrt{x+6} +x}{x^2-4}\\
= \lim\limits_{x\to -2}\dfrac{\left(\sqrt{x+6} +x\right)\left(\sqrt{x+6} -x\right)}{\left(x^2-4\right)\left(\sqrt{x+6} -x\right)}\\
= \lim\limits_{x\to -2}\dfrac{-x^2 + x + 6}{\left(x^2-4\right)\left(\sqrt{x+6} -x\right)}\\
= \lim\limits_{x\to -2}\dfrac{(x+2)(3-x)}{(x-2)(x+2)\left(\sqrt{x+6} -x\right)}\\
= \lim\limits_{x\to -2}\dfrac{3-x}{(x-2)\left(\sqrt{x+6} -x\right)}\\
= \dfrac{3 - (-2)}{\left(-2 - 2\right)\left(\sqrt{-2 + 6} - (-2)\right)}\\
= -\dfrac{5}{16}
\end{array}\)
