Question

lim (x^2-9)((x-3)/abs(x-3)) as x approaches 3 (x->3) using squease theorem

Answer 1

http://www.wolframalpha.com/input/?i=plot+%28x^2-9%29+and+%28x-3%29+and+abs%28x-3%29

Answer 2

http://www.wolframalpha.com/input/?i=lim(x+to+3)+%28x^2-9%29%28x-3%29%2Fabs%28x-3%29

Answer 3

they all squeeze into 0

Answer 4

\[\lim_{x \rightarrow 3^+}\frac{(x^2-9)(x-3)}{|x-3|}=\lim_{x \rightarrow 3+}\frac{(x^2-9)(x-3)}{x-3}=\lim_{x \rightarrow 3^+}(x^2-9)=0\]

Answer 5

split that diff of squares and use it for the other absolute

Answer 6

\[\lim_{x \rightarrow 3^-}\frac{(x^2-9)(x-3)}{|x-3|}=\lim_{x \rightarrow 3^-}\frac{(x^2-9)(x-3)}{-(x-3)}=\lim_{x \rightarrow 3-}-(x^2-9)=0\]

Answer 7

or do we define the abs as (√)^2

Answer 8

since the left limit=right limit
then
\[\lim_{x \rightarrow 3}\frac{(x^2-9)(x-3)}{|x-3|}=0\]