Limits of ratios of polynomials at a point

As $x\to 1$ the numerator and the denominator both tend to $0$, therefore the ratio is in the indeterminate form $\frac{0}{0}$. The numerator is a difference of squares and factors as $(x-1)(x+1)$. Since the limit concerns only the points with $x\ne 1$, the factor $x-1$ cancels and the function coincides, at all these points, with $x+1$. The function $x+1$ is continuous, hence its limit as $x\to 1$ is the value $1+1=2$.

As $x\to 2$ the numerator tends to $4-10+6=0$ and the denominator tends to $4-4=0$, therefore the ratio is in the indeterminate form $\frac{0}{0}$. Both polynomials vanish at $x=2$, hence both contain the factor $x-2$. The numerator has roots $2$ and $3$ and factors as $(x-2)(x-3)$; the denominator is a difference of squares and factors as $(x-2)(x+2)$. Since the limit concerns only the points with $x\ne 2$, the factor $x-2$ cancels. The function $\frac{x-3}{x+2}$ is continuous at $2$, therefore the limit is its value at $2$, that is, $-\frac{1}{4}$.

As $x\to 1$ the numerator tends to $1-1=0$ and the denominator tends to $1-3+2=0$, therefore the ratio is in the indeterminate form $\frac{0}{0}$. The numerator is a difference of cubes and factors as $x^3-1=(x-1)\left(x^2+x+1\right)$. The denominator has roots $1$ and $2$, hence it factors as $(x-2)(x-1)$. Since the limit concerns only the points with $x\ne 1$, the common factor $x-1$ cancels. The function $\frac{x^2+x+1}{x-2}$ is continuous at $1$, therefore the limit is its value at $1$: the numerator equals $3$, the denominator equals $-1$, and the limit is $-3$.