Answer:
[tex]\boxed{\lim_{x\to 3}\dfrac{(3-x)^2}{(x-3)} =0}[/tex]
Step-by-step explanation:
[tex]\lim_{x\to 3}\dfrac{(3-x)^2}{(x-3)}[/tex]
[tex]\lim_{x\to 3}\dfrac{(3-x)^2}{-(-x+3)}[/tex]
[tex]\lim_{x\to 3}\dfrac{(3-x)(3-x)}{-(-x+3)}[/tex]
Cancel the same factor
[tex]\lim_{x\to 3}-(3-x)[/tex]
[tex]\lim_{x\to 3}(-3+x)[/tex]
Now, once as [tex]x \rightarrow 3[/tex],
[tex]\lim_{x\to 3}(-3+x)=0[/tex]
