So, the question is- Find $$ \lim_{x\to 2} \left[\frac{1}{x(x-2)^2}-\frac{1}{x^2-3x+2}\right]$$
What I've tried-
It's quite easy to simplify the limit and get
$$ -\lim_{x\to 2} \left(\frac{x^2-x+1}{(x-2)^2(x-1)x}\right)$$
Which upon putting the value yields-
$$ \left(\frac{3}{0}\right)=-\infty$$
But, the answer in the book is $+\infty$
If I try to find the value of limit using L.H.L and R.H.L, then value comes out to be $-\infty$ and $+\infty$ respectively indicating that limit does not exist in the first place. Where am I going wrong?
It should have been $x^2-3x+1$ instead of $x^2-x+1$
This fixes the problem since:
For $x=2$, $x^2-3x+1$ is negative and $x^2-x+1$ is positive.
Let's look at each factor... You have a negative on the outside already... $x^2-3x+1$ is negative for $x=2$
Now on bottom $(x-2)^2$ is positive from either direction because of the square
$(x-1)$ is positive for $x=2$
$x$ is positive for $x=2$
So you have for $x$ approaches 2 that
$-\frac{x^2-3x+1}{(x-2)^2(x-1)x} \rightarrow - \frac{-}{(+)(+)(+)} \infty$