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Zeroes of a Polynomial

Let fx=x5-1...

Question

Let $f (x) = (x^{5} - 1) (x^{3} + 1), g (x) = (x^{2} - 1) (x^{2} - x + 1)$ and let $h (x)$ be such that $f (x) = g (x) h (x)$ . Then $lim x \to 1 h (x)$ is?

A

0

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B

1

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C

3

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D

5

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E

4

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Solution

The correct option is D $5$

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Similar questions

f (x) = (x^{5} - 1) (x^{3} + 1)

g (x) = (x^{2} - 1) (x^{2} - x + 1)

h (x)

f (x) = g (x) h (x)

lim δ x \to 1 h (x)

f (x) = (x^{5} - 1) (x^{3} + 1), g (x) = (x^{2} - 1) (x^{2} - x + 1)

h (x)

f (x) = g (x) h (x)

lim x \to 1 h (x)

f (x) = [\frac{x}{[x]}], g (x) = [x [\frac{1}{x}]]

h (x) = [\frac{[x]}{x}],

lim x \to 2^{-} f (x) + lim x \to {\frac{1}{2}}^{+} g (x) + lim x \to 2^{+} h (x) =

f (x) = \frac{2}{x - 3}

g (x) = \frac{x - 3}{x + 4}

h (x) = - \frac{2 (2 x + 1)}{x^{2} + x - 12}

lim x \to 3 [f (x) + g (x) + h (x)]

f (x) = x^{2} + \frac{1}{x^{2}}

g (x) = x - \frac{1}{x}, x ϵ R - {- 1, 0, 1}

h (x) = \frac{f (x)}{g (x)}

. Then the local minimum value of

h (x)

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