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x → 3 lim ​ x 2 − 10 x + 21 x 2 + 2 x − 15 ​ = ...

 

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soal harusnya substitusi: . karena hasil substitusi tak tentu, maka limit diselesaikan dengan pemfaktoran:

soal harusnya begin mathsize 14px style limit as x rightwards arrow 3 of fraction numerator x squared plus 2 x minus 15 over denominator x squared minus 10 x plus 21 end fraction equals... space end style 

substitusi:

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow 3 of end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator x squared plus 2 x minus 15 over denominator x squared minus 10 x plus 21 end fraction end cell end table equals fraction numerator 3 squared plus 2 times 3 minus 15 over denominator 3 squared minus 10.3 plus 21 end fraction equals 0 over 0 left parenthesis tak space tentu right parenthesis.

karena hasil substitusi tak tentu, maka limit diselesaikan dengan pemfaktoran:

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 3 of fraction numerator x squared plus 2 x minus 15 over denominator x squared minus 10 x plus 21 end fraction end cell equals cell limit as x rightwards arrow 3 of fraction numerator open parentheses x plus 5 close parentheses open parentheses x minus 3 close parentheses over denominator open parentheses x minus 7 close parentheses open parentheses x minus 3 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 3 of fraction numerator x plus 5 over denominator x minus 7 end fraction end cell row blank equals cell fraction numerator 3 plus 5 over denominator 3 minus 7 end fraction end cell row blank equals cell fraction numerator 8 over denominator negative 4 end fraction end cell row blank equals cell negative 2 end cell end table 

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