For any real number $c$, $\underset{x\to a}{lim}c=c$.

See Properties of Finite Limits for a list of all limit properties.

Since $f(x)=2$ is a constant function, $\underset{x\to 1}{lim}f(x)=2$, which implies that the one-side limit also exists and

If $f(x)$ is a rational function and $a$ is in the domain of $f(x)$, then $\underset{x\to a}{lim}f(x)=f(a)$.

See Properties of Finite Limits for a list of all limit properties.

The function

is a rational function and since $2(-1{)}^2+1\ne 0$ (i.e., the denominator is not equal to zero when $x=-1$), it follows that $-1$ is in the domain of $g$.

Therefore, the limit exists and

The function

is a rational function and since $2-2=0$ (i.e., the denominator equals zero when $x=2$), it follows that $2$ is not in the domain of $h$. Unfortunately, this means that the limit property we used in the previous example does not apply.

When we evaluate the numerator of $h(x)$ at $x=2$, we get

Since the denominator goes to zero but the numerator does not, we can immediately conclude that $\underset{x\to 2}{lim}{\displaystyle \frac{x^2+4x-1}{x-2}}$ does not exist.

Moreover, if we let $x$ approach $2$ from the left (i.e., $x<2$), the values of $(x^2+4x-1)/(x-2)$ will be negative and we can conclude that

And if we let $x$ approach $2$ from the right (i.e., $x>2$), the values of $(x^2+4x-1)/(x-2)$ will be positive and we can conclude that

Even though we can say that the one-sided limits are equal to positive or negative infinity, this still means that the one-sided limits do not exist, and therefore the original limit does not exist either.

If $f(x)$ is a polynomial function, then $\underset{x\to a}{lim}f(x)=f(a)$.

See Properties of Finite Limits for a list of all limit properties.

Since we are approaching 6 from the left, we can assume that $x<6$, which means $h(x)=x^2-4x-12$ and we can apply the limit property of a polynomial.

Since we are approaching 6 from the right, we can assume that $x>6$, which means $h(x)=x^2-5x-6$, and we can apply the limit property of a polynomial.

Since ${\displaystyle \underset{x\to 6^{-}}{lim}h(x)=0}$ and ${\displaystyle \underset{x\to 6^{+}}{lim}h(x)=0}$ (i.e., the two one-sided limits both exist and are equal), we conclude that the limit exists and

Since we are approaching 4 from the left, we can assume that $x<4$, which means $g(x)=\frac{x+5}{3}$ and we can apply the limit property of a polynomial.

Since we are approaching 4 from the right, we can assume that $x>4$, which means $g(x)=6-x^2$, and we can apply the limit property of a polynomial.

Since ${\displaystyle \underset{x\to 4^{-}}{lim}g(x)=3}$ and ${\displaystyle \underset{x\to 4^{+}}{lim}g(x)=-10}$ (i.e., the two one-sided limits both exist but are not equal), we conclude that $\underset{x\to 4}{lim}g(x)$ does not exist.