Objective 3 - Evaluating Limits

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Interpret the notation for limits.

Link to section in online textbook.

Video for evaluating a limit.

Now that we have learned about left- and right-hand limits, we can evaluate the limit of a function at a point.

Evaluating the limit of a function at a point $x=a$ :

$\lim _{x \rightarrow a} (f(x)) = L$

if and only if

$\lim _{x \rightarrow a^{-}} (f(x)) = L = \lim _{x \rightarrow a^{+}} (f(x))$

Note: The limit exists if $L$ is a Real number. The limit can be equal to $\infty$ or $-\infty$ , but we would not say that it exists. If the left- and right-hand limits do not agree, we say the limit does not exist (or DNE for short).

This objective will allow you to practice evaluating the left- and right-hand limits to determine if the limit at a point exists. This would be where you want to practice before the exam.

Answers are either a Real number, $\infty$ , $-\infty$ , or DNE.

Figure 1: Function with a hole at $x=-1$ .

$\lim _{x \rightarrow -\infty } f(x) = \answer {\sage {-Infinity}}$

$\lim _{x \rightarrow -1} g(x) = \answer {3}$

$\lim _{x \rightarrow \infty } f(x) = \answer {\sage {Infinity}}$

Figure 2: Piecewise function to evaluate.

$\lim _{x \rightarrow -\infty } f(x) = \answer {\sage {-Infinity}}$

$\lim _{x \rightarrow -2} f(x) = \answer {3}$

$\lim _{x \rightarrow 1} f(x) = \answer [format=string]{DNE}$

$\lim _{x \rightarrow 3} f(x) = \answer {\sage {-Infinity}}$

$\lim _{x \rightarrow \infty } f(x) = \answer {\sage {Infinity}}$

Figure 3: Piecewise function to evaluate.

$\lim _{x \rightarrow -\infty } f(x) = \answer {\sage {-Infinity}}$

$\lim _{x \rightarrow -2} f(x) = \answer [format=string]{DNE}$

$\lim _{x \rightarrow 0} f(x) = \answer {0}$

$\lim _{x \rightarrow 2} f(x) = \answer [format=string]{DNE}$

$\lim _{x \rightarrow \infty } f(x) = \answer {\sage {Infinity}}$

Figure 4: Piecewise function to evaluate.

$\lim _{x \rightarrow -\infty } f(x) = \answer {\sage {Infinity}}$

$\lim _{x \rightarrow -8} f(x) = \answer {-6}$

$\lim _{x \rightarrow -2} f(x) = \answer [format=string]{DNE}$

$\lim _{x \rightarrow 6} f(x) = \answer [format=string]{DNE}$

$\lim _{x \rightarrow 10} f(x) = \answer {0}$

$\lim _{x \rightarrow \infty } f(x) = \answer {\sage {-Infinity}}$

$f(x) = \frac {\sage {numeratorOnDisplay5}}{\sage {denominatorOnDisplay5}}$

$\lim _{x \rightarrow -\infty } f(x) = \answer {\sage {limitAtNegInfty5}}$

$\lim _{x \rightarrow \sage {value5A} } f(x) = \answer [tolerance=0.01]{\sage {limitAtValue5A}}$

$\lim _{x \rightarrow \sage {value5B} } f(x) = \answer [tolerance=0.01]{\sage {limitAtValue5B}}$

$\lim _{x \rightarrow \infty } f(x) = \answer {\sage {limitAtInfty5}}$

$f(x) = \frac {\sage {numeratorOnDisplay6}}{\sage {denominatorOnDisplay6}}$

$\lim _{x \rightarrow -\infty } f(x) = \answer {\sage {limitAtNegInfty6}}$

$\lim _{x \rightarrow \sage {value6A} } f(x) = \answer [format=string]{DNE}$

$\lim _{x \rightarrow \sage {value6B} } f(x) = \answer [tolerance=0.01]{\sage {limitAtValue6B}}$

$\lim _{x \rightarrow \infty } f(x) = \answer {\sage {limitAtInfty6}}$

$\lim _{x \rightarrow \sage {limitAtEquation6[0]}} \sage {limitAtEquation6[1]} = \answer [tolerance=0.01]{\sage {limitAtEquation6[2]}}$

We can’t plug in the exact value, so we will need to plug in values very near $\sage {limitAtEquation6[0]}$ .

$\lim _{x \rightarrow \sage {eqA}} \frac { \sqrt { x - \sage {eqC} } - \sage {eqB}}{ x - \sage {eqA} } = \answer [tolerance=0.01]{\sage {limitAtEquation7}}$

We can’t plug in the exact value, so we will need to plug in values very near $\sage {eqA}$ .