Evaluate finite and infinite limits in two ways

At $x=1$, the numerator tends to $1+4-1=4$ and denominator to $1+2=3\ne0$, so the limit is $4/3$. The Teacher note also asks for direct substitution without explicitly invoking each limit law: the rational function is continuous at 1, hence the same value follows.

As positive $x$ increases without bound, $1/x$ approaches 0. Thus the limit is 0, even though no finite nonzero $x$ gives $1/x=0$. Infinity is not a number to substitute.

Divide numerator and denominator by $x$: $(3-1/x)/(1+2/x)\to(3-0)/(1+0)=3$. The Teacher note warns that apparent forms such as $\infty/\infty$ or $0/0$ do not give a limit directly; rewrite the expression first. These are indeterminate forms, not evaluated numbers.

Write $(3x-1)/(x+2)=3-7/(x+2)$. The second term tends to 0, so the limit is 3 again. The distinct Teacher note connects this to the horizontal asymptote $y=3$. Keep the original excluded input $x=-2$ when using the rewritten expression.