Q.$\displaystyle\lim_{x\to 2}\left[\frac{x^3-7x+6}{x^3-7x^2+16x-12}\right]$

Numerator: $x^3-7x+6=(x-2)(x+3)(x-1)$ (verified by direct expansion). Denominator: by synthetic division at $x=2$, $x^3-7x^2+16x-12=(x-2)(x^2-5x+6)=(x-2)^2(x-3)$. So the ratio is $\dfrac{(x-2)(x+3)(x-1)}{(x-2)^2(x-3)}=\dfrac{(x+3)(x-1)}{(x-2)(x-3)}$ after cancelling one shared factor. At $x=2$, the numerator $(x+3)(x-1)=5\times1=5\ne0$, but the denominator $(x-2)(x-3)\to0$. So the fraction is unbounded near $x=2$: as $x\to2^-$, $(x-2)<0$ and $(x-3)<0$ so the denominator is a small positive number and the ratio $\to+\infty$; as $x\to2^+$, $(x-2)>0$ and $(x-3)<0$ so the denomin …

Factorize fully (using synthetic division for the denominator), cancel only the factors that genuinely match, and check whether a leftover denominator factor still vanishes at the limit point — if …

Cancelling both copies of $(x-2)$ from the denominator against only one copy in the numerator (an invalid over-cancellation), which would wrongly pr …