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I know there are some limits where you can't to certain substitutions such as $\sin(x)=x$ as $x$ approaches $0$. How do you know when you can or can't do that? I wish I could give you an example because I saw one on this site a few days ago but I can't remember it. (By the way please keep the answers at a calc AB level).

Start wearing purple
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Ovi
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    Are you asking when you can say $\lim_{x \rightarrow x_0} f(x) = f(x_0)$? – Cocopuffs Jun 04 '13 at 20:45
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    For readers not familiar with the AP system, calc AB is a lower-level intro calculus course normally taken in junior or senior year of high school. – Alex Becker Jun 04 '13 at 20:46
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    @Cocopuffs: I had a limit where it was as x approaches 0 of f(x). Then I changed the x with sin(x), and got a wrong result. However, I just realized that I only changed $some$ of the x'es, not all and I guess you have to change all of them to get a right answer. – Ovi Jun 04 '13 at 22:45

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It looks like you might have a slightly weird conception of what a limit is.

For a real valued function, the expression $\lim_{x\to a}f(x)$, when it exists, is a real number. So where you've written "$\sin(x)=x$ as $x$ goes to zero" you haven't properly expressed where the limit is taking place, and the two things on the sides of the equality are not equal at all, except at zero. It would be true to say that $\sin(0)=0$, but I think I agree with Cocopuffs that you probably meant to express $\lim_{x\to 0}\sin(x)=\sin(0)=0$.

This is the general rule (in the context of functions on the reals):

If $f:\Bbb R\to \Bbb R$ is continuous at $a$, then $\lim_{x\to a} f(x)=f(a)$.

rschwieb
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  • I meant that $lim$ $as$ $x$ approaches 0 of $sin(x)=lim$ $as$ $x$ $approaches$ $0$ $of$ $x$. Because the ratio of these limits is $1$, (more clearly $lim$ $as$ $x$ $approaches$ $0$ $of$ $sin(x)/x=1$), in many cases $lim$ $as$ $x$ $approaches$ $0$ $of$ $f(x)=lim$ $as$ $x$ approaches $0$ of $f(sin(x))$. However, in some cases the $limit$ $as$ $x$ $approaches$ $0$ $of$ $f(x)$ does not equal the $limit$ $as$ $x$ $approaches$ $0$ $of$ $f(sin(x))$. My question is, how can we know when making substitutions such as $x=sin(x)$ (of course $as$ $x$ $approaches$ $0$) will not change the answer? – Ovi Jun 04 '13 at 22:28
  • I actually just realized something. I had limit as x approaches 0 of f(x), but I didn't change all of the sin(x) with an x, I just changed some of them, so I didn't have f(sin(x)) – Ovi Jun 04 '13 at 22:41
  • @Ovi Still "$\sin(x)=\lim_{x\to 0}x$" is not right: the right hand side is a real number and the right is a function of $x$. I guess you mean this: $\lim_{x\to 0}\sin(x)=\lim_{x\to 0}x$, which is fine and equates two real numbers. Now in your second sentence, you look like you are describing this: $\lim_{x\to 0}\frac{\sin(x)}{x}=1$ which is also correct, equating two real numbers. I'm almost at my character limit, so another comment is forthcoming... – rschwieb Jun 05 '13 at 10:43
  • @Ovi (continued) If $f(x)$ is any continuous function on the reals, then it is true that $\lim_{x\to 0}f(\sin(x))=f(\lim_{x\to 0}(\sin(x)))=f(0)=\lim_{x\to 0}f(x)$. So if you think they are different, it must be using some discontinuous function $f$. Can you give an example of an $f$ that you have in mind where the "substitution" does not work? – rschwieb Jun 05 '13 at 10:48
  • Well I said in an above comment that I made a mistake; I had lim as x approaches 0 of f(x), but I did not actually find limit as x approaches 0 of f(sin(x)). Instead of replacing all of the x'es with sin(x)'s, I only replaced some, which is why I think I got a wrong answer. An example would be the limit as x approaches 0 of (arcsin(x)-x)/(x^3). I only tried to substitute sin(x) for x in the arcsin function, therefore giving me [arcsin(sin(x))-x]/x^3)=0. However, the limit was 1/6. Now I realize that I would have to substitute all the x'es for sin(x)'s for the lim to be preserved. – Ovi Jun 05 '13 at 11:39
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    @Ovi Good work! – rschwieb Jun 05 '13 at 13:13
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When you want to use approximations such as that $\sin x$ is approximately the same as $x$ for small $x$, then you should consider the limit of their quotient - her the well-known $\lim_{x\to 0}\frac{\sin x}x=1$. A very intuitive way of working with such approximations (though possibly not at your course level) is with the Big-Oh notation: $\sin x = x+O(x^3)$.

Hagen von Eitzen
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  • Well here the limit of their quotient is 1, but is some limit problems if you make this substitution, you get a wrong answer. – Ovi Jun 04 '13 at 22:31
  • I actually just realized something. I had limit as x approaches 0 of f(x), but I didn't change all of the sin(x) with an x, I just changed some of them, so I didn't have f(sin(x)) and I guess not changing all of them does not give you the right answer. – Ovi Jun 04 '13 at 22:42
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The statement "lim as x approaches $0$ of $f(x)/f(sin(x))=1$ is true. Why I was actually getting a wrong answer is because instead of replacing all of the x'es in f(x) with sin(x)'es, I was only replacing some of them. For example, an example would be the limit as $x$ approaches $0$ of $(arcsin(x)-x)/(x^3)$. I only tried to substitute $sin(x)$ for $x$ in the $arcsin$ function, therefore giving me $[arcsin(sin(x))-x]/x^3)=0$. However, the limit was $1/6$, and you do get $1/6$ if you replace $all$ of the $x'es$ with $sin(x)'es.$

Ovi
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