Find the limit as x approaches 0 of (cos^2 x - 4 cos x + 3) / sin^2 x.

Step-by-step solution

1. 1

   Check the form. At $x=0$, $\cos 0=1$ so the numerator is $1-4+3=0$, and $\sin^2 0=0$. The quotient is $\tfrac00$, so some algebra has to happen before substitution can work.

2. 2

   Treat the numerator as a quadratic in $u=\cos x$. It factors over the integers:

   $u^2-4u+3=(u-1)(u-3)\quad\Longrightarrow\quad \cos^2x-4\cos x+3=(\cos x-1)(\cos x-3)$

   The factor $\cos x-1$ is the one that vanishes at $x=0$.

3. 3

   Rewrite the denominator so the same factor appears. Using the Pythagorean identity,

   $\sin^2 x=1-\cos^2 x=(1-\cos x)(1+\cos x)$

   Now both numerator and denominator contain $1-\cos x$ up to a sign — this is what makes the cancellation exact rather than asymptotic.

4. 4

   Cancel. Write $\cos x-1=-(1-\cos x)$:

   $\frac{(\cos x-1)(\cos x-3)}{(1-\cos x)(1+\cos x)}=\frac{-(1-\cos x)(\cos x-3)}{(1-\cos x)(1+\cos x)}=\frac{3-\cos x}{1+\cos x}$

   valid for all $x$ near $0$ with $x\neq 0$, which is all a limit ever needs.

5. 5

   Substitute into the reduced expression. It is now continuous at $x=0$:

   $\lim_{x\to 0}\frac{3-\cos x}{1+\cos x}=\frac{3-1}{1+1}=\frac{2}{2}=1$

6. 6

   Confirm numerically. At $x=10^{-3}$ the original quotient equals $1.0000005$, and at $x=10^{-4}$ it equals $0.99999999$ — both converging to $1$. $\checkmark$