Respuesta :

LammettHash
[tex]\displaystyle\lim_{x\to2}\frac{x^2-x+6}{x+2}[/tex]

Both the numerator and denominator are continuous at [tex]x=2[/tex], which means the quotient rule for limits applies:

[tex]\dfrac{\displaystyle\lim_{x\to2}(x^2-x+6)}{\displaystyle\lim_{x\to2}(x+2)}=\dfrac{2^2-2+6}{2+2}=\dfrac84=2[/tex]

Perhaps you meant to write that [tex]x\to-2[/tex] instead? In that case, you would have

[tex]\displaystyle\lim_{x\to-2}\frac{x^2-x+6}{x+2}=\lim_{x\to-2}\frac{(x+2)(x-3)}{x+2}=\lim_{x\to-2}(x-3)=-2-3=-5[/tex]

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