Question

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

Answer 1

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 "`0/0`" or "`oo/oo`" indefinite symbols.

Generally speaking the rule says that instead of calculating original limit

`lim_{x->x_0}(f(x))/(g(x))`

you can calculate:

`lim_{x->x_0}(f'(x))/(g'(x))`

(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->0}(3(1-cosx))/x=lim_{x->0}(3sinx)/1=0`

Answer 2

A fundamental trigonometric limit is `lim_(xrarr0)sinx/x = 1`.

Explanation:

A second important trigonometric limit is

`lim_(xrarr0) (1-cosx)/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_(xrarr0) (sinx/x * sinx * 1/(1+cosx))`

`= (1) * (0) * (1/(1+1)) = 0`

So for this question,

`lim_(xrarr0) (3(1-cosx))/x = 3lim_(xrarr0)(1-cosx)/x`

`= 3 * 0 = 0`