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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?

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    $\begingroup$ It should be $-\frac{x^2-3x+1}{(x-2)^2(x-1)x}$. $\endgroup$ – Robert Z Jun 2 '17 at 14:24
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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

$-\frac{x^2-3x+1}{(x-2)^2(x-1)x} \rightarrow - \frac{-}{(+)(+)(+)} \infty$

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  • $\begingroup$ Accepted! Try to solving it using LHL & RHL now. $\endgroup$ – Abhishekstudent Jun 2 '17 at 14:25
  • $\begingroup$ That should fix the problem. $\endgroup$ – randomgirl Jun 2 '17 at 14:27
  • $\begingroup$ For $x=2$, $x^2-3x+1$ is negative and $x^2-x+1$ is positive. Shouldn't this mean that value is $-\infty$ $\endgroup$ – Abhishekstudent Jun 2 '17 at 14:35
  • $\begingroup$ Right about the first part... so you have a $-\frac{(-)}{(+)(+)(+)} \infty=+ \infty$ $\endgroup$ – randomgirl Jun 2 '17 at 14:37
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    $\begingroup$ $x^2-3x+1$ is (-) not (+) for $x=2$ $\endgroup$ – randomgirl Jun 2 '17 at 14:40

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