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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

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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