# should i use derivatives?evaluation of limit of function (sin x-sin7x)/8x gives 0/0 if x is approaching to 0. should i use derivatives or another method?

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

To find lim x-->0 [ (sin x-sin7x)/8x], as the initial substitution gives an indeterminate form 0/0, using l'Hopital's rule is the best idea.

Replace the numerator and denominator by their derivatives

=> lim x-->0 [ (cos x - 7*cos 7x)/8]

substitute x = 0

=> (1 - 7*1)/8

=> -6/8

=> -3/4

The required limit is -3/4.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

Since we've get an indetermination, we'll apply l'Hospital rule, rule that use the derivatives of numerator and denominator:

lim (sinx-sin7x)/8x = lim (sinx-sin7x)'/(8x)'

lim (sinx-sin7x)'/8(x)' = lim (cos x- 7cos 7x)/8

We'll substitute x by accumulation point:

lim (cos x- 7cos 7x)/8 = (cos 0- 7cos 7*0)/8

lim (cos x- 7cos 7x)/8 = (1 - 7*1)/8

lim (cos x- 7cos 7x)/8 = -6/8

lim (cos x- 7cos 7x)/8 = -3/4

For x->0, lim (sinx-sin7x)/8x = -3/4