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compute the limit lim (8x-x^2+3)^3 as x goes to 2
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\[\lim_{x\to2}(-x^2+8x+3)^3=(-(2)^2+8(2)+3)^3=?\]
OK SO I JUST SOLVE RIGHT?
Yes
However, when you start seeing things like\[\lim_{x\to2}\frac{x^2+3x+2}{x-2},\]then you have to do some algebraic manipulation.
OK. FOR THE EQUATION ABOVE WOULD IT (-(4)+16+9 ?
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Where did you get that \(9\) from?
I got it from the last part of the equation where it says 3)^3. Or did I do that wrong
yes, that's wrong ^3 is the last step; do what's in the parentheses first
and ^3 does not mean times 3, it means to the third power
so would it be (-4+16+3)^3
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Yes. What does that evaluate to?
-4+16+3= 15 so then I do 15^3 Right? so that would equal 3375
That's right.
Thank you for your help
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