$ \underset{x\to -\infty }{\mathop{\lim }}\,\left( { x ^ 2 }+x-1 \right) $ bằng
A. $ -\infty $
B. $ -2 $
C. $ +\infty $
D. $ 1 $

Cho hàm số $ f\left( x \right)=\left\{ \begin{align} & { x ^ 2 }-3\,\,\,\,\text{khi}\,\,\,\,x\ge 2 \\ & x-1\,\,\,\,\,\,\,\text{khi}\,\,\,\,x < 2 \\ \end{align} \right. $ . Chọn kết quả đúng của $ \underset{x\to 2}{\mathop{\lim }}\,f\left( x \right) $ :
A. $ 0 $ .
B. Không tồn tại.
C. $ -1 $ .
D. $ 1 $ .

$ A=\underset{x\to -2}{\mathop{\lim }}\,\dfrac{x+1}{{ x ^ 2 }+x+4} $ bằng
A. $ -\infty $
B. $ +\infty $
C. $ 1 $
D. $ -\dfrac{1}{6} $

$ B=\underset{x\to \dfrac{\pi } 6 }{\mathop{\lim }}\,\dfrac{{{\sin }^ 2 }2 x -3\cos x }{\tan x} $ bằng
 
A.$ -\infty $
B.$ +\infty $
C.$ 1 $
D.$ \dfrac{3\sqrt{3} } 4 -\dfrac{9}{2} $

$ D=\underset{x\to 1}{\mathop{\lim }}\,\dfrac{\sqrt{3x+1}-2}{\sqrt[3]{3x+1}-2} $ bằng
 
A. $ \dfrac{\sqrt{3} -2}{\sqrt[3] 3 -2} $
B. $ -\dfrac{1}{6} $
C. $ 1 $
D.0

$ A=\underset{x\to 1}{\mathop{\lim }}\,\dfrac{{ x ^ 2 }-x+1}{x+1} $ bằng
A. $ 1 $
B. $ -\infty $
C. $ +\infty $
D. $ \dfrac{1}{2} $

$ \underset{x\to { 2 ^ - }}{\mathop{\lim }}\,\dfrac{{ x ^ 2 }-4}{\sqrt{\left( { x ^ 4 }+1 \right)\left( 2-x \right)}} $ bằng
 
A.$ 1 $
B.$ -\infty $
C.0
D.$ +\infty $

$ \underset{x\to -{ 1 ^ - }}{\mathop{\lim }}\,\dfrac{{ x ^ 2 }+3x+2}{\left| x+1 \right|} $ bằng
A. $ -1 $
B. $ -2 $
C. $ +\infty $
D. $ -\infty $

Mark the letter A, B, C, or D on your answer sheet to indicate the underlined part that needs correction in each of the following questions.
If we had more time last week, we would certainly have finished the project on time.
A.the
B.on time
C.had
D.certainly

Mark the letter A, B, C, or D on your answer sheet to indicate the underlined part that needs correction in the following question.
If one has a special medical condition such as diabetes, epilepsy, or allergy, it is advisable that they carries some kind of identification in order to avoid being given improper medication in an emergency.
A.being
B.carries
C.they
D.has