Remember what L'Hospital's Rule tells us. It says the following.
If [tex]\lim _{x \rightarrow a}\dfrac {f\left( x\right) }{g\left( x\right) } = \dfrac{0}{0} \,\,\textrm{or} \,\,\lim_{x \to a} \dfrac{f(x)}{g(x)} = \dfrac{\pm \infty}{\pm \infty}[/tex]
Then [tex]\lim_{x \to a}\dfrac{f(x)}{g(x)} = \lim_{x \to a} \dfrac{f'(x)}{g'(x)}[/tex]
We know that [tex]\lim _{x\rightarrow 0}\left( x-\sin x\right) =0[/tex] using direct substitution and [tex]\lim _{x\rightarrow 0}\left( x^{3}\right) =0[/tex] through direct substitution as well. Thus, we can find the derivatives of the numerator and denominator and find our limit.
[tex]\dfrac{d}{dx}[x - \sin x] = 1 - \cos x[/tex]
[tex]\dfrac {d}{dx}\left[ x^{3}\right] =3x^{2}[/tex]
Thus, we are now trying to find:
[tex]\lim _{x\rightarrow 0}\dfrac {1-\cos x}{3x^{2}}[/tex]
To do this, we can use direct substitution for both the numerator and denominator.
[tex]\lim _{x\rightarrow 0}(1-\cos x)=1-1=0[/tex]
[tex]\lim _{x\rightarrow 0}\left( 3x^{2}\right) =0[/tex]
Again, we have the following:
[tex]\lim_{x \to 0} \dfrac{1 - \cos x}{3x^2} = \dfrac{0}{0}[/tex]
Thus, we are going to have to use L'Hospital's Rule again.
[tex]\dfrac {d}{dx}\left[ 1-\cos x\right] =\sin x[/tex]
[tex]\dfrac {d}{dx}\left[ 3x^{2}\right] =6x[/tex]
Now, we are trying to find the following:
[tex]\lim _{x\rightarrow 0}\left( \dfrac {\sin x}{6x}\right)[/tex]
Let's find the limits of the numerator and denominator and set them over each other to find the answer to this limit.
[tex]\lim _{x\rightarrow 0}\left( \sin x\right) =0[/tex]
[tex]\lim _{x\rightarrow 0}\left( 6x\right) =0[/tex]
Again, we are going to have to L'Hospital's Rule to find the limit since the answer to our limit is:
[tex]\lim _{x\rightarrow 0}\left( \dfrac {\sin x}{6x}\right) =\dfrac {0}{0}[/tex]
[tex]\dfrac {d}{dx}\left[ \sin x\right] =\cos x[/tex]
[tex]\dfrac {d}{dx}\left( 6x\right) =6[/tex]
Now, we are trying to find:
[tex]\lim _{x\rightarrow 0}\left( \dfrac {\cos x}{6}\right)[/tex]
We are going to use direct substitution to find the limits of the numerator and denominator to find the answer.
[tex]\lim _{x\rightarrow 0}\left( \cos x\right) =1[/tex]
[tex]\lim _{x\rightarrow 0}\left( 6\right) =6[/tex]
Thus, the answer to this limit is:
[tex]\lim _{x\rightarrow 0}\left( \dfrac {\cos x}{6}\right) =\dfrac {1}{6}[/tex]
Our answer is 1/6.
