91Ó°ÊÓ

A limit exists if there is some real number that it is equal to.

We know that the limit expression is given by $\underset{x→c}{\lim }f\left(x\right)=L$.

The limit exists if there is some real number that it is equal to.

Hence, the statement is true.

The limit of$f\left(x\right)$as$x→c$is the value$f\left(c\right)$.

From the limit expression $\underset{x→c}{\lim }f\left(x\right)$, the value is not $f\left(c\right)$, but it is almost $f\left(c\right)$excluding the value at $x=c$.

Hence, the given statement is false.

The limit of $f\left(x\right)$as $x→c$might exist even if the value of$f\left(c\right)$does not.

From the limit expression $\underset{x→c}{\lim }f\left(x\right)$, the limit might exist even if the value $f\left(c\right)$does not exist.

Hence, the statement is true.

The two sided limit of$f\left(x\right)$as$x→c$exists if and only if the left and right limits of$f\left(x\right)$exists as$x→c$.

From the graph, $x$approaches $-1$from the left side the height of the graph approaches $y=1$$\underset{x→-1}{\lim }f\left(x\right)=1$.

If $x$approaches $-1$from the left side the height of the graph approaches $y=1$, role="math" localid="1648018501976" $\underset{x→-1}{\lim }f\left(x\right)=1$.

If $\underset{x→-1}{\lim }f\left(x\right)=\underset{x→-1}{\lim }f\left(x\right)$, then $\underset{x→-1}{\lim }f\left(x\right)=1$.

Hence, the given statement is true.

If the graph of $f$has a vertical asymptote at $x=5$, then$\underset{x→5}{\lim }f\left(x\right)=∞$.

From the graph, $f$has a vertical asymptote, then either $\underset{x→5}{\lim }f\left(x\right)=∞$or $\underset{x→5}{\lim }f\left(x\right)=-∞$.

Hence, the given statement is false.

If$\underset{x→5}{\lim }f\left(x\right)=∞$, then the graph of$f$has a vertical asymptote at$x=5$.

If the limit expression is given by $\underset{x→5}{\lim }f\left(x\right)=∞$or $\underset{x→5}{\lim }f\left(x\right)=-∞$, the graph must have a vertical asymptote at $x=5$.

Hence, the given statement is true.

If$\underset{x→2}{\lim }f\left(x\right)=∞$, then the graph of$f$has a horizontal asymptote at$x=2$.

If the limit expression is given by $\underset{x→2}{\lim }f\left(x\right)=∞$, the graph must have a vertical asymptote at $x=2$and no horizontal asymptote.

Hence, the statement is false.

If $\underset{x→∞}{\lim }f\left(x\right)=2$, then the graph of $f$has a horizontal asymptote at$y=2$.

If the limit expression is given by $\underset{x→∞}{\lim }f\left(x\right)=2$, the graph would have a horizontal asymptote at$y=2$.

Hence, the given statement is true.