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

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#

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#