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?

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

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

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

  • $\begingroup$ 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
  • 1
    $\begingroup$ $x^2-3x+1$ is (-) not (+) for $x=2$ $\endgroup$ – randomgirl Jun 2 '17 at 14:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.