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Ejercicio LΓ­mite SoluciΓ³n lim ⁑ π‘₯ β†’ 2 ( 3 π‘₯ 2 βˆ’ 5 π‘₯ + 2 ) lim xβ†’2 ​ (3x 2 βˆ’5x+2) 4 4 Sustituimos π‘₯ = 2 x=2 en la funciΓ³n lim ⁑ π‘₯ β†’ βˆ’ 1 ( π‘₯ 3 + 2 π‘₯ 2 βˆ’ π‘₯ + 1 ) lim xβ†’βˆ’1 ​ (x 3 +2x 2 βˆ’x+1) 3 3 Sustituimos π‘₯ = βˆ’ 1 x=βˆ’1 en la funciΓ³n lim ⁑ π‘₯ β†’ 0 2 π‘₯ 2 βˆ’ π‘₯ π‘₯ lim xβ†’0 ​ x 2x 2 βˆ’x ​ βˆ’ 1 βˆ’1 Simplificamos la funciΓ³n lim ⁑ π‘₯ β†’ 3 π‘₯ 2 βˆ’ 9 π‘₯ βˆ’ 3 lim xβ†’3 ​ xβˆ’3 x 2 βˆ’9 ​ 6 6 Factorizamos y simplificamos el lΓ­mite lim ⁑ π‘₯ β†’ ∞ 5 π‘₯ 2 + 3 π‘₯ βˆ’ 4 2 π‘₯ 2 βˆ’ π‘₯ + 7 lim xβ†’βˆž ​ 2x 2 βˆ’x+7 5x 2 +3xβˆ’4 ​ 5 2 2 5 ​ Dividimos por π‘₯ 2 x 2 y evaluamos el lΓ­mite