+26376 Lim→3 (x^4 -81)÷(x-3) kann mir jemand zeigen wie ixh daa vereinfachen kann?
\(\small{ \lim \limits_{x\to 3} {\frac{ x^4-81 } {x-3}} = \ ? \qquad | \qquad \small{\text{ Das 3. Binom lautet: }\boxed{~ a^2-b^2 =(a-b)(a+b) ~}} }\)
\(\small{ \begin{array}{rcl} \frac{ x^4-81 } {x-3} &=& \\ &=& \frac{(x^2-9)(x^2+9)}{x-3} \qquad | \qquad x^4-81 = (x^2-9)(x^2+9)\\ &=& \frac{(x-3)(x+3)(x^2+9)}{x-3} \qquad | \qquad x^2-9 = (x-3)(x+3)\\ &=& (x+3)(x^2+9)\\\\ \lim \limits_{x\to 3} {\frac{ x^4-81 } {x-3}} &=& \lim \limits_{x\to 3} {(x+3)(x^2+9)}\\ &=& (3+3)(3^2+9) \\ &=& (3+3)\cdot (9+9)\\ &=& 6 \cdot 18 \\ \mathbf{ \lim \limits_{x\to 3} {\frac{ x^4-81 } {x-3}} } & \mathbf{=} & \mathbf{108} \end{array} }\)
+26376 Lim→3 (x^4 -81)÷(x-3) kann mir jemand zeigen wie ixh daa vereinfachen kann?
\(\small{ \lim \limits_{x\to 3} {\frac{ x^4-81 } {x-3}} = \ ? \qquad | \qquad \small{\text{ Das 3. Binom lautet: }\boxed{~ a^2-b^2 =(a-b)(a+b) ~}} }\)
\(\small{ \begin{array}{rcl} \frac{ x^4-81 } {x-3} &=& \\ &=& \frac{(x^2-9)(x^2+9)}{x-3} \qquad | \qquad x^4-81 = (x^2-9)(x^2+9)\\ &=& \frac{(x-3)(x+3)(x^2+9)}{x-3} \qquad | \qquad x^2-9 = (x-3)(x+3)\\ &=& (x+3)(x^2+9)\\\\ \lim \limits_{x\to 3} {\frac{ x^4-81 } {x-3}} &=& \lim \limits_{x\to 3} {(x+3)(x^2+9)}\\ &=& (3+3)(3^2+9) \\ &=& (3+3)\cdot (9+9)\\ &=& 6 \cdot 18 \\ \mathbf{ \lim \limits_{x\to 3} {\frac{ x^4-81 } {x-3}} } & \mathbf{=} & \mathbf{108} \end{array} }\)
