Mathematics
ShakitaHaston
9

what is limit of f(x) = (x^3-2x^2-9x+4)/(x^2-2x-8) as x approaches 4

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(1) Answers
jackie1977

[latex]\displaystyle\lim_{x\to4}\frac{x^3-2x^2-9x+4}{x^2-2x-8}[/latex] First notice that [latex]x^2-2x-8=(x-4)(x+2)[/latex]. If [latex]x-4[/latex] is not a factor of the numerator, then there is a non-removable discontinuity at [latex]x=4[/latex], and a removable discontinuity otherwise. You have [latex]4^3-2\times4^2-9\times4+4=0[/latex], which means, by the polynomial remainder theorem, that [latex]x-4[/latex] is indeed a linear factor of the numerator. Dividing yields a quotient of [latex]\dfrac{x^3-2x^2-9x+4}{x-4}=x^2+2x-1[/latex] so the limit is [latex]\displaystyle\lim_{x\to4}\frac{x^2+2x-1}{x+2}=\frac{4^2+2\times4-1}{4+2}=\frac{23}6[/latex]

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