lim[x:0,3x(cos7x-1)/(sin7x-7x)] using l'hopital's rule

Question

need the limit using L'hopitals rule

Answer 1

Both numerator and denominator tend to zero as x approaches zero, so the rule applies.

Let f(x)=3x(cos7x-1) and g(x)=sin7x-7x so we have limit x approaches zero of f(x)/g(x).

f'(x)=3x(-7sin7x)+3(cos7x-1)=3cos7x-3-21xsin7x. f''(x)=-21sin7x-147xcos7x-21sin7x=-(42sin7x+147xcos7x).

g'(x)=7cos7x-7. g''(x)=-49sin7x

The rule states that lim f(x)/g(x)= lim f'(x)/g'(x)= lim f''(x)/g''(x) as x approaches zero.

When x is small sinx=x, and sin7x=7x, so -(42sin7x+147xcos7x)=-(294x+147x)=-441x and -49sin7x=-343x, so the quotient becomes 441/343=9/7. The limit is therefore 9/7 or 1.2857. The rule was applied twice when it could be seen that the first derivative limits could not be satisfactorily evaluated.