Why is the limit of (x-1)/(x-3) does not exist and the limit of x-1/(x-3)ex2 for both as x approches 3 from the left and the right
$\displaystyle \displaystyle\lim_{x\nearrow 3}\frac{x-1}{x-3}=\frac{2}{-0}=-\infty$
$\displaystyle \displaystyle\lim_{x\searrow 3}\frac{x-1}{x-3}=\frac{2}{+0}=\infty$
So, the one-sided limits are different and the function has no limit for $\displaystyle x=3$
The second one I think is $\displaystyle \displaystyle\lim_{x\to 3}\frac{x-1}{(x-3)^2}$.
In this case, $\displaystyle \displaystyle\lim_{x\nearrow 3}\frac{x-1}{(x-3)^2}=\frac{2}{+0}=\infty$
and $\displaystyle \displaystyle\lim_{x\searrow 3}\frac{x-1}{(x-3)^2}=\frac{2}{+0}=\infty$
So, the one-sided limits are equal and $\displaystyle \displaystyle\lim_{x\to 3}\frac{x-1}{(x-3)^2}=\infty$.