# How do you find the limit of (3(1-cosx))/x as x->0?

Nov 27, 2016

You can use the de l'Hospital's rule to find this limit. See explanation.

#### Explanation:

The de l'Hospital's Rule is used to calculate limits of expressions leading to "$\frac{0}{0}$" or "$\frac{\infty}{\infty}$" indefinite symbols.

Generally speaking the rule says that instead of calculating original limit

## ${\lim}_{x \to {x}_{0}} \frac{f \left(x\right)}{g \left(x\right)}$

you can calculate:

## ${\lim}_{x \to {x}_{0}} \frac{f ' \left(x\right)}{g ' \left(x\right)}$

(i.e. instead of the limit of quotient of 2 functions you calculate the limit of quotient of their first derivatives)

and those limits will either both be equal or neither of them will exist.

Note: If the limit of quotient of first derivatives is still undefined you can repeat this procedure (calculate the limit of quotient of 2nd derivatives).

Here we have:

${\lim}_{x \to 0} \frac{3 \left(1 - \cos x\right)}{x} = {\lim}_{x \to 0} \frac{3 \sin x}{1} = 0$

Nov 28, 2016

A fundamental trigonometric limit is ${\lim}_{x \rightarrow 0} \sin \frac{x}{x} = 1$.

#### Explanation:

A second important trigonometric limit is

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

The second limit can be proved using the first and continuity of sine and cosine at $0$.

lim_(xrarr0) (1-cosx)/x = lim_(xrarr0) ((1-cosx)(1+cosx))/(x(1+cosx)

 = lim_(xrarr0) sin^2x/(x(1+cosx)

$= {\lim}_{x \rightarrow 0} \left(\sin \frac{x}{x} \cdot \sin x \cdot \frac{1}{1 + \cos x}\right)$

$= \left(1\right) \cdot \left(0\right) \cdot \left(\frac{1}{1 + 1}\right) = 0$

So for this question,

${\lim}_{x \rightarrow 0} \frac{3 \left(1 - \cos x\right)}{x} = 3 {\lim}_{x \rightarrow 0} \frac{1 - \cos x}{x}$

$= 3 \cdot 0 = 0$