3
$\begingroup$

I have this limit:

$$\lim_{x \to 1} \frac{2x^2-x-6}{x(x-1)^3}$$

This can be written as:

$$\lim_{x \to 1^+} \approx \frac{-5}{1\times\mbox{tiny positive}} \to - \infty$$

Why is that? I mean, let's plug in some numbers. $-5/1.0000000001$ is almost $-5$ , the greater the denominator becomes the close the number to $-5$

Can anybody tell me why the book says it goes to -infinity? Thanks a lot

|cite|improve this question
$\endgroup$
  • $\begingroup$ Your post was edited for better readability. You should check, whether everything is ok. You should also choose between $n$ and $x$ - I guess they denote the same thing. (This problem was already in your original post, before the edit.) $\endgroup$ – Martin Sleziak Nov 11 '11 at 14:50
  • $\begingroup$ Believe it or not, text book writers sometimes make mistakes... $\endgroup$ – David Mitra Nov 11 '11 at 14:50
  • $\begingroup$ Indeed, I didn't change any variables, but did try to make it more readable! @Martin $\endgroup$ – The Chaz 2.0 Nov 11 '11 at 14:51
  • $\begingroup$ But more to the point, the limit is indeed equal to $-5$, as you can check for yourself: wolframalpha.com/input/… $\endgroup$ – Martin Sleziak Nov 11 '11 at 14:52
  • 1
    $\begingroup$ I had added a comment that the limit is $-5$. But the function has now changed. There is still the $n$ versus $x$ problem. With the changed function, $\lim_{x\to 1^+}$ is $-\infty$. If we approach from the other direction, limit is $\infty$. So there is no such thing, even in the extended sense, as $\lim_{x \to 1}$. $\endgroup$ – André Nicolas Nov 11 '11 at 15:24
4
$\begingroup$

You didn't choose a tiny positive number! Rather than $1.00000000001$, you should have chosen $0.0000000000001$

One benefit of choosing such tiny decimal numbers is that we can represent them as fractions. So we would get fractions $\frac{1}{1000} , \frac{1}{10000000} , \frac{1}{100000000000} $ ,etc.

But dividing by these fractions is the same as multiplying by $1000, 10000000, 100000000000$, etc. The effect of this is that the fraction grows arbitrarily large.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Thanks a lot. It makes sense now $\endgroup$ – Andrew Nov 11 '11 at 14:55
  • $\begingroup$ Glad to help! . $\endgroup$ – The Chaz 2.0 Nov 11 '11 at 15:15

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.