# How can I evaluate $\lim\limits_{x\to0}\frac{\sin x}{\sqrt{x}}$?

$$f(x)=\lim\limits_{x\to0}\frac{\sin x}{\sqrt{x}}$$

Well, in lot of places I saw that the limit of this function as $x$ tends to zero is zero, but I don't think so. What do you guys think? (I mean limit from both sides.) This will clear what i mean.. Limit from the right exist and is zero, but limit form the left does not exist,, Hence the limit of the function as it tends to 0 does not exist?? or does it? i am just asking that......

• This limit is obviously $0$, what makes you think differently ? – Yves Daoust Dec 11 '16 at 18:07
• @ Yves Daoust, but it is not unique, limit has to be unique,, check it for yourself plz.. – Sarmad Rafique Dec 11 '16 at 18:23
• plz check it now, i have edited it...(sorry my mistake)... – Sarmad Rafique Dec 11 '16 at 18:26
• The limit from the left requires complex numbers; it's doable, but you probably don't want to. – GFauxPas Dec 11 '16 at 18:31
• @wgrenard I wrote my answer before seeing your comment. It is mere coincidence that I chose as my example the same function, $\sqrt x,$ that is used in the question you referred to. I think I prefer the lower-scoring answers of that question over the highest-scoring answer; it is a matter of how clearly the role (or lack of role) of the function's domain is explained. – David K Dec 11 '16 at 19:54

Limits are defined with reference to the domain of the function whose limit is to be taken. This detail may often be left out when the definition of a limit is given in pre-calculus or at the introductory level of calculus. For $f(x)$ to have the limit $L$ as $x$ approaches $x_0,$ you must be able to make all the values of $f(x)$ (except possibly $f(x_0)$ itself) for all values of $x$ that are in the domain of $f$ and in a "neighborhood" of $x_0$ be arbitrarily close to $L,$ simply by making the neighborhood "small enough."

A neighborhood of $x_0$ consists of all points within a certain distance of $x_0$. For the usual measurement of "distance" applied to real numbers, a neighborhood of $x_0$ is just the set of all $x$ such that $x_0-\delta < x < x_0+\delta$ for some positive number $\delta.$ That neighborhood can be divided in two parts, one containing numbers less than $x_0$ and the other containing numbers greater than $x_0.$ When $f$ is defined on both sides of $x_0,$ "left" and "right" limits at $x_0$ can be obtained by looking at what happens in each half of a neighborhood of $x_0.$ If $f$ has a limit at $x_0,$ the "left" and "right" limits will be the same.

But if the domain of $f$ is not all of the real numbers, the domain may not include points on both sides of $x_0$ within very small neighborhoods of $x_0$. For example, consider the square root function from $\mathbb R$ to $\mathbb R.$ Its domain is the interval $[0,\infty).$ No part of the domain is less than $0,$ so when taking the limit at $0$ only the numbers that are in a neighborhood of $0$ and greater than $0$ matter. Therefore it is generally considered correct (except possibly in some elementary courses where "simplified" definitions of limit have been given) to write $$\lim_{x\to0} \sqrt x = 0.$$

That is, when the function $f$ is not defined at all on the "left" side of $x_0,$ but $f$ is defined on the "right" side of $x_0,$ then $\lim_{x\to0} f(x) = \lim_{x\to0^+} f(x).$

Note that if the "left-side" limit of $f$ at $x_0$ is undefined, but the function $f$ itself is defined on the "left side" of $x_0,$ then $f$ does not have a limit at $x_0.$ That is because points on the "left side" of $x_0$ are in the domain of $f$ and therefore must be included in the taking of the limit.

• The first sentence of your answer is the answer!, fantastic!, thanx – Vikram Dec 12 '16 at 14:42
• @ David K,, very well explained thanks... – Sarmad Rafique Dec 12 '16 at 16:12

The limit is $0$. This is because, $\lim_\limits{x\to0}\frac{\sin x}{\sqrt{x}}$ can be written as :$$\lim_\limits{x\to0}\frac{\sqrt{x}\sin x}{x}=\lim_\limits{x\to0}\sqrt{x}\cdot1=0$$ from the well known limit for $\frac{\sin x}{x}$.

EDIT: Since the OP seems to ask for complex values of $x$, the result may even be proved by using power series expansion of $\sin x$ whence the limit is established to be zero.

• plz check it now, i have edited it...(sorry my mistake)... – Sarmad Rafique Dec 11 '16 at 18:25
• ,, That is fine,, but the limit is not unique,, check its graph.. plz – Sarmad Rafique Dec 11 '16 at 18:36
• @SarmadRafique the graph also shows it tending to zero, moreover, the function is defined for $x>0$ – vidyarthi Dec 11 '16 at 18:41
• ,,, and is not defined for x<0,, so limit from the left side???? – Sarmad Rafique Dec 11 '16 at 18:53

If you agree that the limit of $\sqrt{x}$ is $0$ as $x\to0$ (that is, even though its domain is not defined for negative values and there are no values in the neighbourhood of $\epsilon{}$) then you can multiply each side by $\sqrt{x}$ to get $$\lim_\limits{x\to0}\frac{\sqrt{x}\sin{x}}{x}$$ If you know the proof that the limit of $\frac{\sin{x}}{x}$ is 1 (can be easily seen with L'Hopital's rule), then the limit is clearly $0$

• @ Ephemeral,,, That is fine,, but the limit is not unique,, check its graph.. plz – Sarmad Rafique Dec 11 '16 at 18:37

$$\lim_{x\to 0}\frac{\sin x}{\sqrt x}=\lim_{x\to 0}\frac{\sin x}{x}\cdot \sqrt x=\lim_{x\to 0}\frac{\sin x}{x}\cdot \lim_{x\to 0} \sqrt x=1\cdot 0=0$$

It seems to me that your problem is with both, the domain of the function and the definition of limit. So, I will make some comments in this line.

First of all, a function is constituted of three "ingredients":

• a domain;
• a codomain;
• a rule (that, for each element in the domain, assigns a unique element in the codomain).

If we change any of these three ingredients, we obtain a different function which can have different properties (see this for a simple example). So, a function isn't well defined if you only give the algebraic rule (third ingredient above).

However, in the context of real numbers, there is the following convention: to say "the function $f(x)$" without specifying the domain means that the domain is the set of all "acceptable" real numbers $x$ (acceptable in the following sense: for these values of $x$, the value of $f(x)$ is also a real number).

So, in the context of real numbers, the domain of "the function $f(x)=\frac{\sin(x)}{\sqrt{x}}$" is $[0,\infty)$.

Now, the issue of the domain is clear. So, let us we deal with the problem of the limit. In the context of claculus (like in the James Stewart's book) the meaning of $$\lim_{x\to a}f(x)=L\tag{*}$$ is the following: for all $\varepsilon>0$ there exists $\delta >0$ such that $$0<|x-a|<\delta\qquad\Longrightarrow\qquad |f(x)-L|<\varepsilon.$$ This definition is based in the following hypothesis: the function $f$ is defined on some open interval that contains $a$, except possibly at $a$. Well, in your particular example, the point $0$ doesn't satisfy this hypothesis and thus this definition of limit cannot be applied.

In a more general context, for a real function $f$ with domain $D\subset\mathbb{R}$ and a point $a\in D$ that "can be approximated by elements of $D$", the meaning of $(*)$ is the following: for all $\varepsilon>0$ there exists $\delta >0$ such that $$x\in D\text{ and }0<|x-a|<\delta\qquad\Longrightarrow\qquad |f(x)-L|<\varepsilon.$$ According to this definition (which is the right definition for your case), to evaluate $$\lim_{x\to 0}\frac{\sin(x)}{\sqrt{x}}$$ we only have to consider $f(x)$ for $x>0$ (because $x\notin D$ if $x<0$).

Remark 1: the meaning of "$a$ can be approximated by elements of $D$" is the following: for all $r>0$ there exists $x\in D$ such that $x\in (a-r,a+r)$.

Remark 2: the second definition of limit coincides with the first if $a$ satisfies the said hypothesis.

Renark 3: real analysis is what you have to study to make all these things clearer.

• plz check it now, i have edited it...(sorry my mistake)... – Sarmad Rafique Dec 11 '16 at 18:25
• @SarmadRafique I have edited my answer. Please, see it now. – Pedro Dec 12 '16 at 14:16
• ,, very well explained thanks.... – Sarmad Rafique Dec 12 '16 at 16:33