Find the limit as x approaches 0 of sin(9x) divided by cot(12x).

Step-by-step solution

1. 1

   Check whether the limit is actually indeterminate. As $x\to0$, $\cot(12x)=\dfrac{\cos 12x}{\sin 12x}\to\pm\infty$, so the quotient has the form $\dfrac{0}{\infty}$ — which is not an indeterminate form. It is a determinate $0$. Reaching for L'Hôpital here is unnecessary and, since the form is not $0/0$ or $\infty/\infty$, not even valid.

2. 2

   Rewrite the division by a cotangent as a multiplication by a tangent.

   $\frac{\sin(9x)}{\cot(12x)}=\sin(9x)\cdot\tan(12x)=\frac{\sin(9x)\sin(12x)}{\cos(12x)}$

   This single algebraic move removes the infinity from the denominator.

3. 3

   Take the limit of each factor. All three are continuous at $x=0$:

   $\sin(9x)\to0,\qquad\sin(12x)\to0,\qquad\cos(12x)\to1$

4. 4

   Combine by the product and quotient laws.

   $\lim_{x\to0}\frac{\sin(9x)\sin(12x)}{\cos(12x)}=\frac{0\cdot0}{1}=0$

   The denominator is nonzero at the limit point, so the quotient law applies directly.

5. 5

   Confirm the rate as well as the value. Using $\sin u\approx u$, the expression behaves like $9x\cdot12x=108x^2$ near zero. Numerically at $x=10^{-3}$ the quotient is $1.08004\times10^{-4}$ and at $x=10^{-5}$ it is $1.08000\times10^{-8}$ — both match $108x^2$ ✓, so the limit is $0$ and it approaches quadratically.