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Evaluate limx→1(51−x)\lim_{x\to1}\left(\frac{5}{1-x}\right)limx→1(1−x5)
-4
3
does not exist
Evaluate
limx→∞(3x+4)\lim_{x\to\infty}\left(\frac{3}{x+4}\right)limx→∞(x+43)
-2
-1
0
Fill in the blanks
limx→−3(5x+3)\lim_{x\to-3}\left(\frac{5}{x+3}\right)limx→−3(x+35) =
∞
Is this statement true or false?
limx→−1(5x 2+7x+12)\lim_{x\to-1}\left(\frac{5}{x\ ^2+7x+12}\right)limx→−1(x 2+7x+125) is 5
True
False
limx→1(5x 2−1)\lim_{x\to1}\left(\frac{5}{x^{\ 2}-1}\right)limx→1(x 2−15) is
1
undefined
limx→−1((2x−3)(x+1)(x−1)(x+2)(x+1))\lim_{x\to-1}\left(\frac{\left(2x-3\right)\left(x+1\right)\left(x-1\right)}{\left(x+2\right)\left(x+1\right)}\right)limx→−1((x+2)(x+1)(2x−3)(x+1)(x−1)) is ∞
limx→1(5x(x2−1))\lim_{x\to1}\left(\frac{5}{x\left(x^2-1\right)}\right)limx→1(x(x2−1)5)
limx→0(7x(x−1))\lim_{x\to0}\left(\frac{7}{x\left(x-1\right)}\right)limx→0(x(x−1)7)
limx→0(3xx+9)\lim_{x\to0}\left(\frac{3x}{x+9}\right)limx→0(x+93x) =
-9
limx→1(x+1x2−1)\lim_{x\to1}\left(\frac{x+1}{x^2-1}\right)limx→1(x2−1x+1)
It is done.