f(x) = (x⁴+3x³-2x²-x+7) / x-5, when x = 5?
f(x)=(x4+3x3−2x2−x+7)x−5f(x) = \frac{(x^4+3x^3-2x^2-x+7)}{x-5}f(x)=x−5(x4+3x3−2x2−x+7)
At x=5x=5x=5
f(5)=(54+3(5)3−2(5)2−5+7)5−5=∞f(5)=\frac{(5^4+3(5)^3-2(5)^2-5+7)}{5-5}=\inftyf(5)=5−5(54+3(5)3−2(5)2−5+7)=∞
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