Question

How to evaluate the limit

Answer 1

`\displaystyle\lim_(x\to2)(x^2-x+6)/(x+2)`

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


`(\displaystyle\lim_(x\to2)(x^2-x+6))/(\displaystyle\lim_(x\to2)(x+2))=(2^2-2+6)/(2+2)=\frac84=2`

Perhaps you meant to write that
`x\to-2` instead? In that case, you would have


`\displaystyle\lim_(x\to-2)(x^2-x+6)/(x+2)=\lim_(x\to-2)((x+2)(x-3))/(x+2)=\lim_(x\to-2)(x-3)=-2-3=-5`