# What is wrong with this limit reasoning

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

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?

• First order Taylor is not enough because $\infty-\infty$ is indeterminate Apr 29, 2015 at 11:31
• Expand them to $\frac{x-\sin(x)}{x\sin(x)}$ and then use L'Hôpitals theorem often enough. Apr 29, 2015 at 11:33
• @ThorbenKastenholz the first try doesn't fail Apr 29, 2015 at 11:38
• @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 Apr 29, 2015 at 11:41
• Equivalence of functions is not preserved under addition or substraction! It is only preserved under multiplication or division. Apr 29, 2015 at 11:57

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 $\frac{1}{2}$

• I don't understand your second step in both solutions where you bring terms to the numerator. Apr 29, 2015 at 11:57
• How are you bringing the terms to the numerator? Apr 29, 2015 at 11:58
• Just using $\frac{1}{1-x}=1+x+\cdots+x^n+o(x^n)$ Apr 29, 2015 at 12:08
• @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.
– DRF
Apr 29, 2015 at 14:46
• Note that it suffices to use $\sin x = x + O(x^3)$ which gives $\frac 1 {\sin x} = \frac 1 x + O(x)$. Apr 29, 2015 at 19:15

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 $-a$.

• The first try does not fail, its the second one which fails Apr 29, 2015 at 11:36
• both are incorrect justifications. the first fails despite giving the right result. Apr 29, 2015 at 12:51
• @mookid: it is not true that $o(1)\to0$. What happens there is that $x-\sin x=o(x^3)$. Apr 29, 2015 at 14:58
• o(1), by definition, is a function that $\to 0$. Apr 29, 2015 at 18:17

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 $\,F\sim_a G$
• Please check the notation in the first equivalence (assumption) for $f_1$ sim $g_1$. Apr 29, 2015 at 12:51
• Thanks for pointing the typo! Apr 29, 2015 at 12:58

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.