5
$\begingroup$

$$\lim_{x \to 0}\left(\dfrac 1 {\sin x}-\dfrac1 x\right) $$

I solved this problem using the fact that near $x=0$ , $\sin x \cong x$),$(\sin x=x+O(x^2)$ therefore subtracting them results to $0$ which I did verify using other techniques, However, the same reasoning leads to $0$ in solving

$$\lim_{x \to 0}\left(\dfrac 1 {\log (x+1)}-\dfrac1 x\right) $$

while the correct result is $1/2$ although $\log(x+1) \cong x$ when $x$ is near zero$(\log (x+1)=x+O(x^2)$.What is getting wrong here?

$\endgroup$
  • 1
    $\begingroup$ First order Taylor is not enough because $\infty-\infty$ is indeterminate $\endgroup$ – marwalix Apr 29 '15 at 11:31
  • $\begingroup$ Expand them to $\frac{x-\sin(x)}{x\sin(x)}$ and then use L'Hôpitals theorem often enough. $\endgroup$ – ThorbenK Apr 29 '15 at 11:33
  • 1
    $\begingroup$ @ThorbenKastenholz the first try doesn't fail $\endgroup$ – curious Apr 29 '15 at 11:38
  • 1
    $\begingroup$ @ThorbenKastenholz i know, how to solve this one using L.Hopital, what I want to know is that why the method gives a correct result in the first case and an erroneous one in the second $\endgroup$ – curious Apr 29 '15 at 11:41
  • 1
    $\begingroup$ Equivalence of functions is not preserved under addition or substraction! It is only preserved under multiplication or division. $\endgroup$ – Bernard Apr 29 '15 at 11:57
13
$\begingroup$

We will be using the following three Taylor expansions: $\sin{x}=x-\frac{x^3}{6}+O(x^5)$, $\log(1+x)=x-\frac{x^2}{2}+O(x^3)$ and $\frac{1}{1-x}=1+x+O(x^2)$

$$\begin{align}\frac{1}{\sin{x}}&=\frac{1}{x-\frac{x^3}{6}+O(x^5)}\\&=\frac{1}{x(1-x^2/6+O(x^4))}\\&=\frac{1}{x}(1+\frac{x^2}{6}+O(x^4))\\&=\frac{1}{x}+\frac{x}{6}+O(x^3)\end{align}$$

And the limit in this case is zero. While

$$\begin{align}\frac{1}{\log(1+x)}&=\frac{1}{x-\frac{x^2}{2}+O(x^3)}\\&=\frac{1}{x(1-x/2+O(x^2))}\\&=\frac{1}{x}(1+\frac{x}{2}+O(x^2))\\&=\frac{1}{x}+\frac{1}{2}+O(x)\end{align}$$

And the limit is $\frac{1}{2}$

$\endgroup$
  • 2
    $\begingroup$ I don't understand your second step in both solutions where you bring terms to the numerator. $\endgroup$ – curious Apr 29 '15 at 11:57
  • $\begingroup$ How are you bringing the terms to the numerator? $\endgroup$ – GFauxPas Apr 29 '15 at 11:58
  • 6
    $\begingroup$ Just using $\frac{1}{1-x}=1+x+\cdots+x^n+o(x^n)$ $\endgroup$ – marwalix Apr 29 '15 at 12:08
  • 4
    $\begingroup$ @mawalix +1 and +1 for the usage of the $\frac{1}{1-x}$ expansion. But consider adding it into the answer instead of just a comment. $\endgroup$ – DRF Apr 29 '15 at 14:46
  • 1
    $\begingroup$ Note that it suffices to use $\sin x = x + O(x^3)$ which gives $\frac 1 {\sin x} = \frac 1 x + O(x)$. $\endgroup$ – yoann Apr 29 '15 at 19:15
2
$\begingroup$

The first try fails because the limit of both terms is $\infty$. You need the stronger result $$ \sin x = x + o(x^2) $$

and then you get $$ \frac 1{\sin x} - \frac 1x = \frac{x - \sin x}{x\sin x} = \frac{o(x^2)}{x\sin x} = o(1)\to 0 $$

More generally if $$ f(x) = x + a x^2 + o(x^2) $$(important case: $f$ is twice differentiable around 0 with $a = f''(0)/2$) then the limit is $-a$.

$\endgroup$
  • 1
    $\begingroup$ The first try does not fail, its the second one which fails $\endgroup$ – curious Apr 29 '15 at 11:36
  • 1
    $\begingroup$ both are incorrect justifications. the first fails despite giving the right result. $\endgroup$ – mookid Apr 29 '15 at 12:51
  • $\begingroup$ @mookid: it is not true that $o(1)\to0$. What happens there is that $x-\sin x=o(x^3)$. $\endgroup$ – Martin Argerami Apr 29 '15 at 14:58
  • $\begingroup$ o(1), by definition, is a function that $\to 0$. $\endgroup$ – mookid Apr 29 '15 at 18:17
2
$\begingroup$

Your error comes from the fact that equivalence of functions is not compatible with addition and substraction.

Counter-example:

We have $x+x^2\sim_0 x$, $\,-x\sim_0 -x+x^3$, but $$(x+x^2)-x=x^2 \not\sim_0 x+(-x+x^3)=x^3. $$

While equivalence is a powerful tool for computing limits as it frees from irrelevant computational details, it must be used cautiously:

Main rules for computing with equivalence:

Let $f,g$ functions defined in a neighbourhood of $a \in \overline{\mathbf R}$. If $f\sim_a g $, $f_1\sim_a g_1$, then:

  • $\dfrac1f\sim_a \dfrac1g$
  • for all $n\in\mathbf N$, $\,f^n\sim_a g^n$
  • $ff_1\sim_a g\,g_1$
  • $\dfrac f{f_1}\sim_a \dfrac g{g_1}$
  • if $\,F'=f, \enspace G'=g\,$ and $\,F(a)=G(a)=0$ then $\,F\sim_a G$
$\endgroup$
  • $\begingroup$ Please check the notation in the first equivalence (assumption) for $f_1$ sim $g_1$. $\endgroup$ – CiaPan Apr 29 '15 at 12:51
  • $\begingroup$ Thanks for pointing the typo! $\endgroup$ – Bernard Apr 29 '15 at 12:58
0
$\begingroup$

This is one of the most common problems associated with the use of symbol $\sim$ (or here $\cong$ as used by OP).

The error comes because of a lack of proper understanding of the definition of symbol $\sim$. The proper definition of this symbol is as follows:

If $\lim\limits_{x \to a}\dfrac{f(x)}{g(x)} = 1$ then we write $f(x) \sim g(x)$ as $x \to a$.

From the above definition it should be clear enough (although many students almost always fail to get it) that this definition is in a multiplicative context. Thus if $f(x) \sim g(x)$ as $x \to a$ then we can replace $f(x)$ by $g(x)$ while calculating the limit of an expression containing $f(x)$ provided $f(x)$ is in a multiplicative context in that expression. Here "multiplicative context" means that the expression containing $f(x)$ must be of type $f(x)h(x)$ (or like $h(x)/f(x)$). And then we can as well calculate the limit of $g(x)h(x)$ and get answer.

This is because $$\lim_{x \to a}f(x)h(x) = \lim_{x \to a}\frac{f(x)}{g(x)}\cdot g(x)h(x) = \lim_{x \to a}1\cdot g(x)h(x)$$

It is in this manner that we replace $f(x)$ by $g(x)$ when $f(x) \sim g(x)$. When calculating limits one must know exactly which rules of limits are used in each step (even when they are being used implicitly).

In the current context $\log(1 + x) \sim x$ as $x \to 0$, but expression $$\left(\frac{1}{\log(1 + x)} - \frac{1}{x}\right)$$ is not of the form $\log(1 + x)\cdot \text{(some function)}$. Well you may write the expression as $$\frac{1}{\log(1 + x)}\left(1 - \frac{\log(1 + x)}{x}\right)$$ and the first factor definitely be replaced by $1/x$ to get $$\frac{1}{x}\left(1 - \frac{\log(1 + x)}{x}\right)$$

Note: Many people try to complicate the things by adding the talk of "order". One should not attach any more meaning to the statement $f(x) \sim g(x)$ than is provided by its definition. Trying to attach more meaning to this statement like "$f(x)$ is almost equal to $g(x)$ to first order of smallness" is only going to add confusion and not make it more intuitive.

$\endgroup$

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.