Question

I need some help with this practice problem in calc

Answer 1

So we have to find the limit when x tends to 3 of:

`\lim _(x\to3)(x^3-9x)/(3x^2-6x-9)`

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:

`\lim _(x\to c)(f(x))/(g(x))=\lim _(x\to c)(f^(\prime)(x))/(g^(\prime)(x))`

In this case, our function is composed of two functions. Let's see if we can apply the rule to them:

`\begin{gathered} f(x)=x^3-9x \\ g(x)=3x^2-6x-9 \end{gathered}`

Let's check their limits when x tends to 3:

`\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}`

They are polynomials so they are differentiable in any real interval and g'(x) is not always zero:

`g^(\prime)(x)=(d)/(dx)(3x^2-6x-9)=2\cdot3x-6=6x-6`

So they meet the conditions and we can apply the L'Hôpital's rule:

`\begin{gathered} \lim _(x\to3)(x^3-9x)/(3x^2-6x-9)=\lim _(x\to3)((d)/(dx)(x^3-9x))/((d)/(dx)(3x^2-6x-9))=\lim _(x\to3)(3x^2-9)/(6x-6)=(3\cdot3^2-9)/(6\cdot3-6) \\ \lim _(x\to3)(x^3-9x)/(3x^2-6x-9)=(27-9)/(18-6)=(18)/(12)=(3)/(2) \end{gathered}`

So the limit we were looking for is 3/2. Then the answer is the fourth option.