`a)lim_{x->3^[-]} [-4x]/[x-3]=+oo`
`b)lim_{x->3^[+]} [-4x]/[x-3]=-oo`
Đúng 2
Bình luận (0)
Các câu hỏi tương tự
tính
a) \(\lim\limits_{x\rightarrow3^-}\dfrac{-4x}{x-3}\)
b) \(\lim\limits_{x\rightarrow3^+}\dfrac{-4x}{x-3}\)
tính giới hạn
a) \(\lim\limits_{x\rightarrow3}\dfrac{\sqrt{2x+10}-4}{3x-9}\)
b) \(\lim\limits_{x\rightarrow7}\dfrac{\sqrt{4x+8}-6}{x^2-9x+14}\)
c) \(\lim\limits_{x\rightarrow5}\dfrac{x^2-8x+15}{2x^2-9x-5}\)
4. Tính giới hạn \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x^2+1}-x-1}{2x^2-x}_{ }\)
5. Tính giới hạn:
a) \(\lim\limits_{x\rightarrow2}\dfrac{x-2}{x^2-4}_{ }\)
b) \(\lim\limits_{x\rightarrow3^-}\dfrac{x+3}{x-3}_{ }\)
tính giới hạn
a) \(\lim\limits_{x\rightarrow3}\dfrac{x^2-9}{x^2-5x+6}\)
b) \(\lim\limits_{x\rightarrow5}\dfrac{x^2-5x}{x-5}\)
c) \(\lim\limits_{x\rightarrow-3}\dfrac{x^2-3x}{2x^2+9x+9}\)
cho \(f\left(x\right)=\left\{{}\begin{matrix}x^2-3\\x+3\end{matrix}\right.\) \(x\ge3\);\(x< 3\)
a) tính \(\lim\limits_{x\rightarrow3^+}f\left(x\right)=?\)
\(\lim\limits_{x\rightarrow3^-}f\left(x\right)=?\)
b) tính \(\lim\limits_{x\rightarrow3}f\left(x\right)\) nếu có
Cho em hỏi cách bấm CASIO giải trắc nghiệm với ạ
\(\lim\limits_{x\rightarrow1}\)\(\dfrac{x^2+2x-3}{2x^2-x-1}\)
\(\lim\limits_{x\rightarrow3^-}\dfrac{\left|1-3x\right|}{3-x}\)
11) \(\lim\limits_{x->1}\) \(\dfrac{3_{\sqrt{4x-1}-\sqrt{4x-3}}}{x-1}\)
11) \(\lim\limits_{x->4}\dfrac{4x-1}{x^2-8x+16}\)
12) \(\lim\limits_{x->2}\)\(\dfrac{4-x^2}{x^3-8}\)
13) \(\lim\limits_{x->+\infty}\left(3_{\sqrt{x^3+4x^2}-x}\right)\)
tính giới hạn
a) \(\lim\limits_{x\rightarrow4}\dfrac{\sqrt{2x+8}-4}{x-4}\)
b) \(\lim\limits_{x\rightarrow2}\dfrac{x^2-4}{\sqrt{4x+1}-3}\)
c) \(\lim\limits_{x\rightarrow2}\dfrac{x-2}{2-\sqrt{x+2}}\)
tính giới hạn
a) \(\lim\limits_{x\rightarrow+\infty}\dfrac{5x^2+x^3+5}{4x^3+1}\)
b) \(\lim\limits_{x\rightarrow-\infty}\dfrac{2x^2-x+1}{x^3+x-2x^2}\)
c) \(\lim\limits_{x\rightarrow-\infty}\dfrac{2x^2-x+1}{x^3+x-2x^2}\)