Los limm^2-m+1/m^2+m+1=1 op

$$\lim \frac { m ^ { 2 } - m + 1 } { m ^ { 2 } + m + 1 } = 1$$

Oplossen voor l

$$l=\frac{1}{Im(\frac{1}{m^{2}+m+1})\left(Re(m^{2})-Re(m)+1\right)+Re(\frac{1}{m^{2}+m+1})\left(Im(m^{2})-Im(m)\right)}$$

$$Im(\frac{1}{m^{2}+m+1})\left(Re(m^{2})-Re(m)+1\right)+Re(\frac{1}{m^{2}+m+1})\left(Im(m^{2})-Im(m)\right)\neq 0\text{ and }m\neq \frac{-1+\sqrt{3}i}{2}\text{ and }m\neq \frac{-\sqrt{3}i-1}{2}$$

Stappen voor oplossen lineaire vergelijking

De vergelijking heeft de standaardvorm.

$$\left(Im(\frac{1}{m^{2}+m+1})\left(Re(m^{2})-Re(m)+1\right)+Re(\frac{1}{m^{2}+m+1})\left(Im(m^{2})-Im(m)\right)\right)l=1$$

Deel beide zijden van de vergelijking door \left(Re(m^{2})-Re(m)+1\right)Im(\left(m^{2}+m+1\right)^{-1})+\left(Im(m^{2})-Im(m)\right)Re(\left(m^{2}+m+1\right)^{-1}).

$$\frac{\left(Im(\frac{1}{m^{2}+m+1})\left(Re(m^{2})-Re(m)+1\right)+Re(\frac{1}{m^{2}+m+1})\left(Im(m^{2})-Im(m)\right)\right)l}{Im(\frac{1}{m^{2}+m+1})\left(Re(m^{2})-Re(m)+1\right)+Re(\frac{1}{m^{2}+m+1})\left(Im(m^{2})-Im(m)\right)}=\frac{1}{Im(\frac{1}{m^{2}+m+1})\left(Re(m^{2})-Re(m)+1\right)+Re(\frac{1}{m^{2}+m+1})\left(Im(m^{2})-Im(m)\right)}$$

Delen door \left(Re(m^{2})-Re(m)+1\right)Im(\left(m^{2}+m+1\right)^{-1})+\left(Im(m^{2})-Im(m)\right)Re(\left(m^{2}+m+1\right)^{-1}) maakt de vermenigvuldiging met \left(Re(m^{2})-Re(m)+1\right)Im(\left(m^{2}+m+1\right)^{-1})+\left(Im(m^{2})-Im(m)\right)Re(\left(m^{2}+m+1\right)^{-1}) ongedaan.

$$l=\frac{1}{Im(\frac{1}{m^{2}+m+1})\left(Re(m^{2})-Re(m)+1\right)+Re(\frac{1}{m^{2}+m+1})\left(Im(m^{2})-Im(m)\right)}$$