Consider the function $f\left(x\right)=\frac{1}{7-x}$f(x)=17−x.
Complete the following table of values, in which $x<7$x<7.
$x$x | $5$5 | $6$6 | $6.9$6.9 | $6.99$6.99 |
$f\left(x\right)$f(x) | $\editable{}$ | $\editable{}$ | $\editable{}$ | $\editable{}$ |
Complete the following table of values, in which $x>7$x>7.
$x$x | $9$9 | $8$8 | $7.1$7.1 | $7.01$7.01 |
$f\left(x\right)$f(x) | $\editable{}$ | $\editable{}$ | $\editable{}$ | $\editable{}$ |
What is the limit of $f\left(x\right)$f(x) as the value of $x$x approaches $7$7?
The limit does not exist.
The limit is $0$0.
The limit is $\infty$∞.
The limit is $-\infty$−∞.
Consider the function $f\left(x\right)=\frac{1}{x+3}$f(x)=1x+3.
Consider the function $f\left(x\right)=\frac{x^2+7x+10}{x+5}$f(x)=x2+7x+10x+5.