Question

Câu 21. Tính giới hạn : $\underset{ x\to -2}{\mathop{\lim}} \dfrac{x-1+\sqrt{2 x^2+1}}{4-x^2}$.

Answer 1

\(\lim\limits_{x\rightarrow-2}=\dfrac{x-1+\sqrt{2x^2+1}}{4-x^2}\)

\(=\lim\limits_{x\rightarrow-2}=\dfrac{\left[\left(x-1\right)+\sqrt{2x^2+1}\right]\left[\left(x-1\right)-\sqrt{2x^2+1}\right]}{\left(4-x^2\right)\left[\left(x-1\right)-\sqrt{2x^2+1}\right]}\)

\(=\lim\limits_{x\rightarrow-2}\dfrac{\left(x-1\right)^2-\left(2x^2+1\right)}{\left(4-x^2\right)\left[\left(x-1\right)-\sqrt{2x^2+1}\right]}\)

\(=\lim\limits_{x\rightarrow-2}\dfrac{x^2-2x+1-2x^2-1}{\left(4-x^2\right)\left[\left(x-1\right)-\sqrt{2x^2+1}\right]}\)

\(=\lim\limits_{x\rightarrow-2}\dfrac{-x^2-2x}{\left(4-x^2\right)\left[\left(x-1\right)-\sqrt{2x^2+1}\right]}\)

\(=\lim\limits_{x\rightarrow-2}=-\dfrac{x}{\left(2-x\right)\left(x-1-\sqrt{2x^2+1}\right)}\)

\(=-\dfrac{1}{12}\)

Answer 2

\(\overset{lim}{x\rightarrow-2}\dfrac{x-1+\sqrt{2x^2+1}}{4-x^2}\) = \(\overset{lim}{x\rightarrow-2}\dfrac{x^2-2x+1-2x^2-1}{\left(4-x^2\right)\left(x-1-\sqrt{2x^2+1}\right)}\)

                                           = \(\overset{lim}{x\rightarrow-2}\dfrac{-x^2-2x}{\left(4-x^2\right)\left(x-1-\sqrt{2x^2+1}\right)}\)

                                          = \(\overset{lim}{x\rightarrow-2}\dfrac{-x\left(x+2\right)}{-\left(x-2\right)\left(x+2\right)\left(x-1-\sqrt{2x^2+1}\right)}\)

                                          = \(\overset{lim}{x\rightarrow-2}\dfrac{x}{\left(x-2\right)\left(x-1-\sqrt{2x^2+1}\right)}\)

                                          = \(\dfrac{-2}{\left(-2-2\right)\left[-2-1-\sqrt{2.\left(-2\right)^2+1}\right]}=-\dfrac{1}{12}\)

Answer 3

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