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belka [17]
3 years ago
6

Lim x-1 x³-2x²+3x-2/(2x^4-3x+1)

Mathematics
1 answer:
UkoKoshka [18]3 years ago
4 0

Since the limit becomes the undetermined form

\displaystyle \lim_{x\to 1} \dfrac{x^3-2x^2+3x-2}{2x^4-3x+1} \to \dfrac{0}{0}

it means that both polynomials have a root at x=1. So, we can fact both numerator and denominator:

x^3-2x^2+3x-2 = (x-1)(x^2-x+2)

2x^4-3x+1 = (x-1)(2x^3+2x^2+2x-1)

So, the fraction becomes

\dfrac{(x-1)(x^2-x+2)}{(x-1)(2x^3+2x^2+2x-1)} = \dfrac{x^2-x+2}{2x^3+2x^2+2x-1}

Now, as x approaches 1, you have no problems anymore:

\displaystyle \lim_{x\to 1} \dfrac{x^2-x+2}{2x^3+2x^2+2x-1} \to \dfrac{2}{5}

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