# When I can't convert a piece of a limit in a notable limit?

When I solve limits I often convert a block in a notable limit multiplying and dividing it by the same quantity. For example: $$\lim_{x \to 0} \frac{e^{\sin(x)}-1}{x} = \lim_{x \to 0} \frac{e^{\frac{\sin(x)}{x}x}-1}{x} = \lim_{x \to 0} \frac{e^{x}-1}{x} = 1$$

I've noticed that I can't always apply this. For example: $$\lim_{x \to 0} \frac{x^3+9x-9 \tan(x)}{-8x^3} = \frac{1}{4}\neq \lim_{x \to 0} \frac{x^3+9x-9 \frac{\tan(x)}{x}x}{-8x^3} = \lim_{x \to 0} \frac{x^3+9x-9x}{-8x^3}= -\frac{1}{8}$$

What's wrong in that passage? What rule am I violating?

Thanks

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Technically, you can't so blithely make the change in the first example either. It has to be justified in terms of the order of vanishing of $\sin(x)/x$ vs. that of $x$. –  Arturo Magidin May 3 '12 at 17:59
To repeat what Arturo Magidin wrote, your first derivation was incorrect, but the answer happened to be right. You can argue this way if you have full knowledge of the fine-grained behaviour of the function near $x=0$, usually through knowledge of the first few relevant terms of the series expansion. –  André Nicolas May 3 '12 at 18:04
@AndréNicolas Let me know if there is anything sloppy going on with my answer. I guess it is clear, but it is rather a delicate topic. –  Pedro Tamaroff May 3 '12 at 19:24

Intuitively, the problem is that $\tan x$ has a term in the Taylor series proportional to $x^3$ which you are ignoring, but the terms of interest are of order $x^3$. You can ignore the terms of order $x^5$ and up because they will contribute $0$ in the limit.

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I never noticed this connection with Taylor series before... nice! –  rschwieb May 3 '12 at 19:19
@rschwieb: many of the class problems in limits are designed to cancel out the high order terms in the Taylor series. As the trig expansions are every other order, you can get deep easily. I remember seeing one that required terms of order $x^8$ –  Ross Millikan May 3 '12 at 19:34
I have not yet been exposed to the phenomenon. I only barely remember the expansions from undergraduate calculus. Fringe benefits of participation here :) –  rschwieb May 3 '12 at 19:36
@rschwieb: Things like $\lim_{x \to 0}\frac {\tan x - \sin x}{x^3}$ –  Ross Millikan May 3 '12 at 19:38

I'll try expand on Ross' answer.

Functions like $\sin x$, $\tan x$ and $\exp x$ can be approximated by Taylor polynomials near $x=0$ (and usually around any $x=a$ that makes sense - note the delicate case for $\tan x$).

What can be enlightening to consider in the subtitutions you are making is the order of the approximation relative to the greatest power of $x$. We say a function $\alpha(x)$ is an infinitesimal when $x \to a$ if

$$\lim_{x \to a} \alpha(x) = 0$$

We say two infinitesimals are of the same order if

$$\lim \frac \alpha \beta =A\neq 0$$

If it is the case $A=1$, we say $\alpha$ and $\beta$ are equivalent infinitesimals and we write

$$\alpha \sim \beta$$

Given two infinitesimals $\alpha$ and $\beta$ in $x \to a$, we say $\beta$ is of greater order than $\alpha$ if

$$\lim \frac \alpha \beta =0$$

This means, $\alpha$ is "smaller" at $x =a$ than $\beta$.

Likewise, we say that $\beta$ is an infinitesimal of order $n$ respect to $\alpha$ if $\alpha^n$ and $\beta$ are of the same order.

$$\lim \frac \beta {\alpha^n} =A$$

As you note, there are some notable limits, such as

$$e^x -1\sim x$$

$$\sin x \sim x$$

$$\log (x+1) \sim x$$

These can all be proven from the Taylor series definition's or some other arifice. But the Taylor series tells us more. In fact, Taylor series tells us about the degree of approximation we obtain when we approximate a function, such as $\sin x$, by a certain $n$ degree polynomial. And the result is that

If $T_n^a(f,x)$ is the Taylor polynomial of $f$, of $n$th degree around $x=a$, then

$$f(x) -T_n^a(f,x)=o[(x-a)^n] \text{ ; } x \to a$$

Read: $f(x) -T_n^a(f,x)$ is an infinitesimal of smaller order than $(x-a)^n$ when $x \to a$, or $f(x) -T_n^a(f,x)$ is small-$o$ of $(x-a)^n$ when $x \to a$.

The idea, as you see, is that

$$\lim \frac{f(x) -T_n^a(f,x)}{(x-a)^n} =0$$

The importance of this is that it tells us the error made by approximating $f$ by it's $n$th degree polynomial, is small in comparison to $(x-a)^n$. When we swap between functinons and their approximation when calculating limits, we have to make sure the approximation is of order smaller than the greatest degree of $x$ we see, otherwise, the error will not go to zero when we take the limit! In your example, you are implicitly noting that

$$\tan x+o(x) = x$$

You state that

$$\mathop {\lim }\limits_{x \to 0} \frac{{{x^3} + 9x - 9\tan (x)}}{{ - 8{x^3}}} = \frac{1}{4}$$

\eqalign{ & \mathop {\lim }\limits_{x \to 0} \frac{{{x^3} + 9x - 9\left( {x + o\left( x \right)} \right)}}{{ - 8{x^3}}} = \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{{x^3} + 9x - 9x - 9o\left( x \right)}}{{ - 8{x^3}}} = \cr & = - \frac{1}{8} + \mathop {\lim }\limits_{x \to 0} \frac{9}{8}\frac{{o\left( x \right)}}{{{x^3}}} \cr}

And there is an error that doesn't go away, because it would only go away if the denomitor had been $x$, or a greater infinitesimal (1 for example). But is is $x^3$. To fix this, we use

$$\tan x = x + \frac{{{x^3}}}{3} + o\left( {{x^3}} \right)$$

This gives

\eqalign{ & \mathop {\lim }\limits_{x \to 0} \frac{{{x^3} + 9x - 9\left( {x + \frac{{{x^3}}}{3} + o\left( {{x^3}} \right)} \right)}}{{ - 8{x^3}}} \cr & \mathop {\lim }\limits_{x \to 0} \frac{{{x^3} + 9x - 9x - 3{x^3} - 9o\left( {{x^3}} \right)}}{{ - 8{x^3}}} \cr & \mathop {\lim }\limits_{x \to 0} \frac{{ - 2{x^3}}}{{ - 8{x^3}}} + \mathop {\lim }\limits_{x \to 0} \frac{9}{8}\frac{{o\left( {{x^3}} \right)}}{{{x^3}}} = \frac{1}{4} \cr}

which is correct.

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Thank you very much for this detailed explanation! But 1/4 was correct, in fact you forgot a 3 before the second $x^3$ ;) –  xonya May 3 '12 at 19:37
@user771431 Whoops! So silly. –  Pedro Tamaroff May 3 '12 at 19:51
After 19th century mathematicians expended so much effort to banish infinitesimals! (What you called infinitesimals are not the infinitesimals of non-standard analysis.) –  André Nicolas May 3 '12 at 21:34
@AndréNicolas I know. They are just the name given to $$\lim_{x\to a} f(x) \to 0$$. –  Pedro Tamaroff May 3 '12 at 21:48
It is ordinarily best to use standard names and notations. –  André Nicolas May 3 '12 at 21:52

The $x$'s always have to go "together" to the limit, or else you could do silly things like this:

$\lim\limits_{x\rightarrow 0^+}\frac{x}{x}=?!=\lim\limits_{x\rightarrow 0^+}\frac{0}{x}=0$ when the real limit is actually 1.

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