Limits at a point with the indeterminate form zero over zero

Substituting $x=2$, numerator and denominator vanish, so the limit is in the indeterminate form $\frac{0}{0}$. A polynomial that vanishes at $2$ is divisible by $x-2$, therefore both are factored, writing the numerator as a difference of cubes and the denominator as a difference of squares. The factor $x-2$ cancels, because in the limit only $x\ne 2$ matters. The remaining function is continuous at $2$, so the limit is computed by substitution and equals $3$.

For $x=1$ both trinomials equal $0$, so the form is $\frac{0}{0}$ and $x=1$ is a root of both. Each trinomial factors as the product of $x-1$ and a first-degree factor, which is found from the other root, that is $-3$ for the numerator and $-\frac{1}{2}$ for the denominator, therefore $2x^2-x-1=2(x-1)\left(x+\frac{1}{2}\right)=(x-1)(2x+1)$. After cancelling, we substitute $x=1$ and obtain $\frac{4}{3}$.

Here too substitution gives $\frac{0}{0}$, but the numerator contains a root and does not factor like a polynomial. We rationalize by multiplying numerator and denominator by $\sqrt{x}+2$, and so the numerator becomes the difference of squares $x-4$, which cancels with the denominator. Alternatively, one can observe that $x-4=(\sqrt{x}-2)(\sqrt{x}+2)$ for $x\ge 0$. The remaining function is continuous at $4$ and the limit equals $\frac{1}{4}$.