Answer:
[tex]\lim_{x \to 0} \frac{3sinx-3x}{7x^3}=-\frac{1}{14}[/tex]
Step-by-step explanation:
The limit is:
[tex]\lim_{x \to 0} \frac{3sinx-3x}{7x^3}=\frac{0}{0}[/tex]
so, you have an indeterminate result. By using the l'Hôpital's rule you have:
[tex]\lim_{x \to 0} \frac{a(x)}{b(x)}= \lim_{x \to 0} \frac{a'(x)}{b'(x)}[/tex]
by replacing, and applying repeatedly you obtain:
[tex]\lim_{x \to 0} \frac{3sinx-3x}{7x^3}= \lim_{x \to 0}\frac{3cosx-3}{21x^2}= \lim_{x \to 0}\frac{-3sinx}{42x}= \lim_{x \to 0}\frac{-3cosx}{42}\\\\ \lim_{x \to 0} \frac{3sinx-3x}{7x^3}=\frac{-3cos0}{42}=-\frac{1}{14}[/tex]
hence, the limit of the function is -1/14
