Evaluate the limit limx→π2(9sec(x)18+7tan(x))\lim_{x\to{\frac{\pi }{2}}}\left(\frac{9\sec\left(x\right)}{18+7\tan\left(x\right)}\right)limx→2π(18+7tan(x)9sec(x)) by replacing all occurrences of xxx by π2\frac{\pi }{2}2π
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limx→−1(12x3+12x2x4−x2)\lim_{x\to-1}\left(\frac{12x^3+12x^2}{x^4-x^2}\right)x→−1lim(x4−x212x3+12x2)
limx→1(x3−14x3−x−3)\lim_{x\to1}\left(\frac{x^3-1}{4x^3-x-3}\right)x→1lim(4x3−x−3x3−1)
limx→8(x−8x2−64)\lim_{x\to8}\left(\frac{x-8}{x^2-64}\right)x→8lim(x2−64x−8)
limx→3(x3−27x2−9)\lim_{x\to3}\left(\frac{x^3-27}{x^2-9}\right)x→3lim(x2−9x3−27)
limx→4(2−x4−x)\lim_{x\to4}\left(\frac{2-\sqrt{x}}{4-x}\right)x→4lim(4−x2−x)
limx→2(x4−16x−2)\lim_{x\to2}\left(\frac{x^4-16}{x-2}\right)x→2lim(x−2x4−16)
limx→−4(x+4x2−16)\lim_{x\to-4}\left(\frac{x+4}{x^2-16}\right)x→−4lim(x2−16x+4)
Find limits of functions at a specific point by directly plugging the value into the function.
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