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高二数学 | 高级数学
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12.2 Limit of Functions 函数的极限


Evaluate the following limits
计算下列各极限

[1]

\lim\limits_{h\rightarrow0}\frac{(2+h)^2-4}{h}

Answer:
4
[2]

\lim\limits_{x\rightarrow5}\frac{x-5}{2x^2-9x-5}

Answer:
\frac{1}{11}
[3]

\lim\limits_{x\rightarrow-5}\frac{(x+1)^3+(x-3)^2}{x(x+5)}

Answer:
-6\frac{2}{5}
[4]

\lim\limits_{x\rightarrow-2}\frac{x^3+8}{x+2}

Answer:
12
[5]

\lim\limits_{t\rightarrow9}\frac{9-t}{3-\sqrt t}

Answer:
6
[6]

\lim\limits_{t\rightarrow0}\frac{\sqrt{2-t}-\sqrt2}{t}

Answer:
-\frac{\sqrt2}{4}
[7]

\lim\limits_{x\rightarrow1}\frac{x^3-2x^2-5x+6}{x^2-1}

Answer:
-3
[8]

\lim\limits_{x\rightarrow1}(\frac{1}{x-1}-\frac{2}{x^2-1})

Answer:
\frac{1}{2}
[9]

\lim\limits_{t\rightarrow0}(\frac{1}{t\sqrt{1+t}}-\frac{1}{t})

Answer:
-\frac{1}{2}
[10]

\lim\limits_{x\rightarrow9}\frac{x^2-81}{\sqrt x-3}

Answer:
108
[11]

\lim\limits_{x\rightarrow1}\frac{2-\sqrt{x+3}}{x^2-1}

Answer:
-\frac{1}{8}
[12]

\lim\limits_{x\rightarrow2}\frac{\frac{1}{x}-\frac{1}{2}}{x-2}

Answer:
-\frac{1}{4}
[13]

\lim\limits_{x\rightarrow1}\frac{2-\sqrt[4]{16x}}{2-\sqrt[3]{8x}}

Answer:
\frac{3}{4}
[14]

\lim\limits_{x\rightarrow-2}\frac{|x+1|-1}{x+2}

Answer:
-1
[15]

\lim\limits_{x\rightarrow0}\frac{x^2}{x^2+|x|}

Answer:
0
[16]

\lim\limits_{x\rightarrow2}\frac{\sqrt{3x-2}-\sqrt{x+2}}{\sqrt{x-1}-\sqrt{3-x}}

Answer:
\frac{1}{2}
[17]

If g(x) = \begin{cases} ax^2+3 , x \le 1 \\ x+2a , x>1\end{cases}, find the value of a if \lim\limits_{x\rightarrow1}g(x)

若g(x) = \begin{cases} ax^2+3 , x \le 1 \\ x+2a , x>1\end{cases} , 求 a 的值若 \lim\limits_{x\rightarrow1}g(x)

Answer:
2
[18]

Given that \lim\limits_{x\rightarrow2}\frac{x^3+px^2+qx-2}{x-2}=5, find the values of p and q.

已知 \lim\limits_{x\rightarrow2}\frac{x^3+px^2+qx-2}{x-2}=5 , 求 p 和 q 的值.

Answer:
–2, 1
[19]

Find the constants a and b such that \lim\limits_{x\rightarrow0}\frac{\sqrt{ax+b}-2}{x}=1

求常数a 和 b使得 \lim\limits_{x\rightarrow0}\frac{\sqrt{ax+b}-2}{x}=1

Answer:
4 , -4