Answer:
[tex]\dfrac{-1}{6}[/tex]
Step-by-step explanation:
Given the limit of a function expressed as [tex]\lim_{ x\to \ 0} \dfrac{sin(x)-tan(x)}{x^3}[/tex], to evaluate the following steps must be carried out.
Step 1: substitute x = 0 into the function
[tex]= \dfrac{sin(0)-tan(0)}{0^3}\\= \frac{0}{0} (indeterminate)[/tex]
Step 2: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the function
[tex]= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ sin(x)-tan(x)]}{\frac{d}{dx} (x^3)}\\= \lim_{ x\to \ 0} \dfrac{cos(x)-sec^2(x)}{3x^2}\\[/tex]
Step 3: substitute x = 0 into the resulting function
[tex]= \dfrac{cos(0)-sec^2(0)}{3(0)^2}\\= \frac{1-1}{0}\\= \frac{0}{0} (ind)[/tex]
Step 4: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 2
[tex]= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ cos(x)-sec^2(x)]}{\frac{d}{dx} (3x^2)}\\= \lim_{ x\to \ 0} \dfrac{-sin(x)-2sec^2(x)tan(x)}{6x}\\[/tex]
[tex]= \dfrac{-sin(0)-2sec^2(0)tan(0)}{6(0)}\\= \frac{0}{0} (ind)[/tex]
Step 6: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 4
[tex]= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ -sin(x)-2sec^2(x)tan(x)]}{\frac{d}{dx} (6x)}\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^2(x)sec^2(x)+2sec^2(x)tan(x)tan(x)]}{6}\\\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^4(x)+2sec^2(x)tan^2(x)]}{6}\\[/tex]
Step 7: substitute x = 0 into the resulting function in step 6
[tex]= \dfrac{[ -cos(0)-2(sec^4(0)+2sec^2(0)tan^2(0)]}{6}\\\\= \dfrac{-1-2(0)}{6} \\= \dfrac{-1}{6}[/tex]
Hence the limit of the function [tex]\lim_{ x\to \ 0} \dfrac{sin(x)-tan(x)}{x^3} \ is \ \dfrac{-1}{6}[/tex].
