Question

Use​ l'Hôpital's Rule to find the following limit. ModifyingBelow lim With x right arrow 0StartFraction 3 sine (x )minus 3 x Over 7 x cubed EndFraction ModifyingBelow lim With x right arrow 0StartFraction 3 sine (x )minus 3 x Over 7 x cubed EndFraction equals nothing ​(Type an exact​ answer.)

Answer 1

`\lim_(x \to 0) (3sinx-3x)/(7x^3)=-(1)/(14)`

Explanation:

The limit is:

`\lim_(x \to 0) (3sinx-3x)/(7x^3)=(0)/(0)`

so, you have an indeterminate result. By using the l'Hôpital's rule you have:

`\lim_(x \to 0) (a(x))/(b(x))= \lim_(x \to 0) (a'(x))/(b'(x))`

by replacing, and applying repeatedly you obtain:

`\lim_(x \to 0) (3sinx-3x)/(7x^3)= \lim_(x \to 0)(3cosx-3)/(21x^2)= \lim_(x \to 0)(-3sinx)/(42x)= \lim_(x \to 0)(-3cosx)/(42)\\\\ \lim_(x \to 0) (3sinx-3x)/(7x^3)=(-3cos0)/(42)=-(1)/(14)`

hence, the limit of the function is -1/14