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Suppose that $ F $, $ G $, and $ Q $ are polynomials and$$ \frac{F(x)}{Q(x)} = \frac{G(x)}{Q(x)} $$for all x except when $ Q(x) = 0 $. Prove that $ F(x) = G(x) $ for all $ x $. [Hint: Use continuity.]

$$\begin{aligned}F(a) &=\lim _{x \rightarrow a} F(x) & &[\text { by continuity of } F] \\&=\lim _{x \rightarrow a} G(x) & &[\text { whenever } Q(x) \neq 0] \\&=G(a) & &[\text { by continuity of } G]\end{aligned}$$

Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 4

Integration of Rational Functions by Partial Fractions

Integration Techniques

Missouri State University

Oregon State University

Harvey Mudd College

University of Michigan - Ann Arbor

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F G and Cure polynomial. Is that satisfied? This equation here for all exes except when Q A zero. And then we'd like to go ahead and show that F equals G for all ex. So here, I guess. Case. One. No way should consider, but it's kind of a silly case here. This is a Q is just the zero polynomial. So let me go ahead and put this bar here. So this is his cues. It's not like you've X's hero for some X. It's zero for every ex. So if this is true, then this we automatically have a people's G because we'LL never find an example in which F is not equal to G for some X satisfying. Q. Thanks not equal to zero because this can't happen because he was always zero. So in that case, it's vacuous. Lee Truth. There's really nothing to show then. The second case is Kyu is only zero for some X. So for finally, many exes, a polynomial could have on ly finally, many zeroes. So it's either always zero are on ly zero finally many times. So now if, for example, if Q of A is not equal to zero Then we can use this equation up here and we have enough of a over Cuba Guia que es. Since Cuba is non zero, we can just cancel these and then we see f of a equals jia come So a satisfies this equation here at those a values f equals gene. Now suppose that queue of a does equal zero This is what we'Ll have to use the continuity and that's what the hell is suggesting. So if ave this's weaken right, this is a limit as X approaches a f of x this's by continuity of f f is continuous because it's a polynomial This is given So then we can, right? This is the limit His ex approaches a of genomex So this is of course, when that satisfies this equation here can be zero in the denominator. No. So again, we have to make this assumption here. If we want this last equation now, using continuity of G, we can write this last expression Is she of eh using continuity of G. So this is because ji is continuous. But here's what we shown we shown for any hey whether queue of a zero or not. We always have equals G. So this shows that vehicle's G for all ex values, and that's your final answer.

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