limn→∞((1)⋅arctan (n)arctan (n+1))\lim_{n\to\infty}\left(\frac{\left(1\right)\cdot\arctan\:\left(n\right)}{\arctan\:\left(n+1\right)}\right)n→∞lim(arctan(n+1)(1)⋅arctan(n))
limx→5(x2−25x2+2x−25)\lim_{x\to5}\left(\frac{x^2-25}{x^2+2x-25}\right)x→5lim(x2+2x−25x2−25)
−3a2b3a3-3a^2b^3a^3−3a2b3a3
25x2−125x^2-125x2−1
−20x(+10):(−5):(+8)-20x\left(+10\right):\left(-5\right):\left(+8\right)−20x(+10):(−5):(+8)
x2−14x+13x^2-14x+13x2−14x+13
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