So we have to find the limit when x tends to 3 of:
[tex]\lim _{x\to3}\frac{x^3-9x}{3x^2-6x-9}[/tex]For this purpose we can use the L'Hôpital's rule. It states that when the limits of f(x) and g(x) are equal to 0 or ±∞, f and g are differentiable in an interval around a number x=c, and g'(x) is not zero then we have:
[tex]\lim _{x\to c}\frac{f(x)}{g(x)}=\lim _{x\to c}\frac{f^{\prime}(x)}{g^{\prime}(x)}[/tex]In this case, our function is composed of two functions. Let's see if we can apply the rule to them:
[tex]\begin{gathered} f(x)=x^3-9x \\ g(x)=3x^2-6x-9 \end{gathered}[/tex]Let's check their limits when x tends to 3:
[tex]\begin{gathered} \lim _{x\to3}f(x)=\lim _{x\to3}x^3-9x=3^3-9\cdot3=0 \\ \lim _{x\to3}g(x)=\lim _{x\to3}3x^2-6x-9=3\cdot3^2-6\cdot3-9=0 \end{gathered}[/tex]They are polynomials so they are differentiable in any real interval and g'(x) is not always zero:
[tex]g^{\prime}(x)=\frac{d}{dx}(3x^2-6x-9)=2\cdot3x-6=6x-6[/tex]So they meet the conditions and we can apply the L'Hôpital's rule:
[tex]\begin{gathered} \lim _{x\to3}\frac{x^3-9x}{3x^2-6x-9}=\lim _{x\to3}\frac{\frac{d}{dx}(x^3-9x)}{\frac{d}{dx}(3x^2-6x-9)}=\lim _{x\to3}\frac{3x^2-9}{6x-6}=\frac{3\cdot3^2-9}{6\cdot3-6} \\ \lim _{x\to3}\frac{x^3-9x}{3x^2-6x-9}=\frac{27-9}{18-6}=\frac{18}{12}=\frac{3}{2} \end{gathered}[/tex]So the limit we were looking for is 3/2. Then the answer is the fourth option.
