# How do you find the limit of x/sinx as x approaches 0?

Mar 3, 2016

$1$

#### Explanation:

Let $f \left(x\right) = \frac{x}{\sin} x$

$\implies f ' \left(x\right) = {\lim}_{x \to 0} \frac{x}{\sin} x$

$\implies f ' \left(x\right) = {\lim}_{x \to 0} \frac{1}{\sin \frac{x}{x}} = \frac{{\lim}_{x \to 0} 1}{{\lim}_{x \to 0} \left(\sin \frac{x}{x}\right)} = \frac{1}{1} = 1$

Oct 5, 2016

$1$

#### Explanation:

We need to know the important trigonometric limit:

${\lim}_{x \rightarrow 0} \sin \frac{x}{x} = 1$

So, we see that:

${\lim}_{x \rightarrow 0} \frac{x}{\sin} x = {\lim}_{x \rightarrow 0} {\left(\sin \frac{x}{x}\right)}^{-} 1 = {\left({\lim}_{x \rightarrow 0} \sin \frac{x}{x}\right)}^{-} 1 = {1}^{-} 1 = 1$

Oct 27, 2016

${\lim}_{x \rightarrow 0} \frac{x}{\sin x} = 1$

#### Explanation:

If we try to calculate the limit directly, we can see that is an indeterminate form:

lim_{x rarr 0} x/{sin x} = 0/{sin 0} = 0/0 ?

To solve it, we can apply the L'Hôpital's rule:

Given two functions $f$ and $g$ differentiable at $x = a$, it holds that:

If ${\lim}_{x \rightarrow a} \frac{f \left(x\right)}{g \left(x\right)} = \frac{0}{0}$ or ${\lim}_{x \rightarrow a} \frac{f \left(x\right)}{g \left(x\right)} = \frac{\infty}{\infty}$ then:

${\lim}_{x \rightarrow a} \frac{f \left(x\right)}{g \left(x\right)} = {\lim}_{x \rightarrow a} \frac{f ' \left(x\right)}{g ' \left(x\right)}$

So we have:

${\lim}_{x \rightarrow 0} \frac{x}{\sin x} = {\lim}_{x \rightarrow 0} \frac{1}{\cos x} = \frac{1}{\cos 0} = \frac{1}{1} = 1$.

NOTE
The question was posted in "Determining Limits Algebraically" , so the use of L'Hôpital's rule is NOT a suitable method to solve the problem. Therefore this solution is invalid.

This limit can not be solved using only algebraic concepts as the function $\sin \left(x\right)$ is not an algebraic function. We may use a Taylor series to approximate $\sin \left(x\right)$ by a polynomial and thus use a linear or quadratic approximation that would calculate the limit, but also emerge the issue of derivatives.

The answer above that uses the limit ${\lim}_{x \rightarrow 0} \frac{\sin x}{x}$ also is invalid (using the criteria indicated by the note) because this limit cited needs also L'Hôpital's rule to be improved. It is not correct to say that is an important limit and that is why we must know if we can not prove it in the context that is intended for use.

I think the person who wrote the first note confuses the term "algebraic" in the expression "Determining Limits algebraically" in the true meaning of the word algebraic, regardless of that expression usually means "Determination of limits analytically", ie, by calculation and not graphically, numerically (using approximations) or intuitive.

Oct 28, 2016

Use the squeeze theorem;
${\lim}_{x \to 0} \frac{x}{\sin} x = 1$

#### Explanation:

We can use the squeeze theorem (or sandwich theorem), which states that if $g \left(x\right) \le f \left(x\right) \le h \left(x\right)$ in an interval around $c$ then ${\lim}_{x \to c} g \left(x\right) \le {\lim}_{x \to c} f \left(x\right) \le {\lim}_{x \to c} h \left(x\right)$ (providing the limits exist), and that if ${\lim}_{x \to c} g \left(x\right) = l = {\lim}_{x \to c} h \left(x\right)$ then ${\lim}_{x \to c} f \left(x\right) = l$

So the challenge is to find upper and lower bounds for $\frac{x}{\sin} x$

Consider:

We have:
{Area of $\Delta$ KOA} $\le$ {Area of Sector KOA} $\le$ {Area of $\Delta$ LOA}

$\therefore \sin \frac{x}{2} \le \frac{x}{2} \le \tan \frac{x}{2}$

Multiplying by $\frac{2}{\sin} x$ gives:
$1 \le \frac{x}{\sin} x \le \frac{1}{\cos} x$

$\therefore {\lim}_{x \to 0} 1 \le {\lim}_{x \to 0} \frac{x}{\sin} x \le {\lim}_{x \to 0} \frac{1}{\cos} x$

$\therefore 1 \le {\lim}_{x \to 0} \frac{x}{\sin} x \le 1$

Hence, by the squeeze theorem, ${\lim}_{x \to 0} \frac{x}{\sin} x = 1$