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I have to find the limit of following

$$\lim_{x \to 0}\left(\frac{1}{x} - \frac{1}{x^2}\right)$$

I have no idea how to start this one off. How would I do it?

Do I just substitute the $0$? It doesn't look that easy and simple. The answer says it's negative infinity.

Please show me a solution without graphing(unless for better explanation).

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    $\begingroup$ For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$ – Tacet Nov 29 '14 at 1:56
  • $\begingroup$ thanks for the useful links $\endgroup$ – didgocks Nov 29 '14 at 1:57
  • $\begingroup$ have you evaluated the function $\frac{1}{x} - \frac{1}{x^2}$ for small values od $x$ like $x = \pm 0.1, \pm 0.01, \cdots?$ $\endgroup$ – abel Dec 3 '14 at 15:53
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$$\lim_{x\to 0}\frac{x-1}{x^2} = \lim_{x\to 0}\frac{-1}{x^2} = -\infty$$

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    $\begingroup$ Your first equality above can be pretty suspicious or even incomprehensible to a beginner $\endgroup$ – Timbuc Nov 29 '14 at 2:02
  • $\begingroup$ How did you exactly get that steps to the answer? $\endgroup$ – didgocks Nov 29 '14 at 2:19
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    $\begingroup$ This answer evaluates the limit in the numerator first; that is $\lim_{x\rightarrow 0}x-1=-1$, so it's safe to replace the numerator by the constant $-1$. (This works as long as the numerator doesn't go to $0$). $\endgroup$ – Milo Brandt Nov 29 '14 at 3:35
  • $\begingroup$ @Meelo, I think many know that. The question is whether the OP knows it and even more important: whether he can justify it. Besides this, it is in general a mistake, and even a big one, to evaluate part of a mathematical expression and not the rest and, anyway, this requires a proof. $\endgroup$ – Timbuc Nov 29 '14 at 11:06
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    $\begingroup$ Your first step is wrong (however the equality between the two things is correct). There is no general theorem to replace a sub-expression $x$ occurring in a complicated expression $(x - 1)/x^{2}$ with its limit $0$ in the given circumstances. $\endgroup$ – Paramanand Singh Mar 30 '16 at 4:01
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An idea. We take $\;x\;$ very close to zero, say $\;|x|<10^{-4}\;$ :

$$x>0:\;\;\frac1x-\frac1{x^2}=\frac{x-1}{x^2}<\frac{-\frac12}{x^2}=-\frac1{2x^2}$$

and now you only have to show the rightmost expresion is unbounded below, which I think is pretty easy.

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As written, the limit is in the so-called “indeterminate form $\infty-\infty$”), so we want to rewrite it in another way to start with: $$ \frac{1}{x}-\frac{1}{x^2}=\frac{x-1}{x^2} $$ It's not restrictive to work under the assumption that $-1<x<1$; thus $|x|<1$ and $|x|^2<|x|$, that is to say $$ \frac{1}{x^2}>\frac{1}{|x|} $$ Since $\lim_{x\to0}(x-1)=-1$, we can restrict ourselves to an interval around $0$ where $x-1<-1/2$, so $$ \frac{x-1}{x^2}<\frac{-1/2}{x^2}<-\frac{1}{2|x|} $$ Since $$ \lim_{x\to0}-\frac{1}{2|x|}=-\infty $$ we are done.


However, this can be stated in greater generality; if you know that

  1. $\displaystyle\lim_{x\to a}f(x)=l>0$ (possibly $l=\infty$)
  2. $\displaystyle\lim_{x\to a}g(x)=0$
  3. $g(x)>0$ in a neighborhood of $a$ ($a$ excluded)

then $$ \lim_{x\to a}\frac{f(x)}{g(x)}=\infty $$

Note that the limit can also be for $x\to a^+$ or $x\to a^-$; changing into $l<0$ or $g(x)<0$ is easy with the “rule of signs”.

The proof is just the same as before: since $\lim_{x\to a}f(x)=l>0$, we can restrict ourselves to a (punctured) neighborhood of $a$ where $f(x)>k$ for some $k>0$. Then, since $\lim_{x\to a}g(x)=0$, for any $M>0$ we can choose $\delta>0$ so that, for $0<|x-a|<\delta$, $|g(x)-0|<k/M$. Thus, as we can also assume $g(x)>0$, $1/g(x)>M/k$ and $$ \frac{f(x)}{g(x)}>k\frac{M}{k}=M $$ This is exactly proving that $\lim_{x\to a}f(x)/g(x)=\infty$.

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  • $\begingroup$ +1, This is probably the best answer to this question. $\endgroup$ – k170 Nov 25 '19 at 0:23
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For $f(x)=\frac{1}{x}-\frac{1}{x^2}$, if you want to know what $\lim_{x \rightarrow 0}f(0)$ is, try looking at $f(\frac{1}{10}),f(\frac{-3}{100})$ and selected similar values. That will start you off and give you an idea what the answer might be (or why the book says that the answer is what it is.) Once you have decided that the answer is $- \infty$ you might want to prove it. Graphing is, in some sense, a way to look at $f(x)$ for lots of values at the same time.

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A useful thing to do would be to make the substitution $u=\frac{1}x$. Then, this becomes $$\lim_{u\rightarrow\infty}u-u^2$$ (or the analogous limit to $-\infty$) but $u$ grows much more slowly than $u^2$, the expression in the limit must decrease without bound - in particular, since $u^2>2u$ if $u>2$, we get that $u-u^2<-u$ if $u$ is at least $2$, so the limit is bounded above by $-u$ which goes to $-\infty$.

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    $\begingroup$ It's pretty hard to justify this substitution without using harder machinery than the problem needs directly. $\endgroup$ – djechlin Nov 29 '14 at 3:40
  • $\begingroup$ I suppose that's true - but if you wanted to avoid substitution, the important note here is that $\frac{1}{x^2}=(\frac{1}x)^2$ and $|\frac{1}x\rightarrow| \infty$ when $x\rightarrow 0$. You can use that directly in the limit, but I thin the substitution makes it clearer. $\endgroup$ – Milo Brandt Nov 29 '14 at 3:47
  • $\begingroup$ I think this substitution is a good idea, but the explanation about what grows faster and what slower may be lacking some kind of formality this kind of beginning exercises usually ask. Nevertheless, this could probably be left to the OP to complete. $\endgroup$ – Timbuc Nov 29 '14 at 11:08
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Here's another way to prove the following statement $$\lim\limits_{x \to 0}\left(\frac{1}{x} - \frac{1}{x^2}\right)=-\infty$$ Which is equivalent to the following $$\forall N\gt0,\exists\delta\gt 0:0\lt\left|x\right|\lt\delta\Rightarrow\frac{1}{x}-\frac{1}{x^2}\lt -N$$ So whenever $0\lt\left|x\right|\lt\delta$, we have $$\frac{1}{x}-\frac{1}{x^2}=\frac{x-1}{x^2}\lt\frac{\delta-1}{x^2}$$ Here we can make $\left|x\right|$ arbitrarily small by presupposing a bound of $\frac12$. We can then use this bound to potentially get a smaller $\delta$ later. Assuming that $|x|\lt\frac12$, we have $$x^2\lt\frac14\Rightarrow-\frac{1}{x^2}\lt-4$$ Which implies that $$\frac{\delta-1}{x^2}=-\frac{1-\delta}{x^2}\lt -4(1-\delta)=-N$$ Now we can set $\delta$ to the smallest of these two bounds $$\delta =\min\left(\frac12, 1-\frac{N}{4}\right)$$ Putting it all together $$0\lt\left|x\right|\lt\delta\Rightarrow\frac{1}{x}-\frac{1}{x^2}\lt -4(1-\delta)=-N$$ Therefore $$\lim\limits_{x \to 0}\left(\frac{1}{x} - \frac{1}{x^2}\right)=-\infty$$

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