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Explain why if \(\lim _{x \rightarrow a} f(x)\) exist, then so do \(\lim _{x \rightarrow a^{-}} f(x)\) and \(\lim _{x \rightarrow a^{+}} f(x)\).
If \(\lim _{x \rightarrow a} f(x)\) exists, say \(\lim _{x \rightarrow a} f(x) = L\) then the \(f(x)\) values approach \(L\) as \(x \rightarrow a\) from both directions. Hence both one-sided limits exist. In particular, \(\lim _{x \rightarrow a^{-}} f(x) = L\) and
\(\lim _{x \rightarrow a^{+}} f(x) = L\).
Find an instance (There are a few examples to be found in Example limitfromgraphex …) where \(\lim _{x \rightarrow a^{-}} f(x)\) and \(\lim _{x \rightarrow a^{+}} f(x)\) both exist but \(\lim _{x \rightarrow a} f(x)\) does
not.
In Example limitfromgraphex, \(\lim _{x \rightarrow -1^{-}} f(x) = 0\) and \(\lim _{x \rightarrow -1^{+}} f(x) = 4\) both exist but \(\lim _{x \rightarrow -1} f(x)\) does not because the two one-sided limits are not equal.
Consider the complete graph of the function \(g\) below. Use the graph to find the indicated values.
If a limit fails to exist, state that is the case or use the symbols ‘\(\infty \)’ or ‘\(-\infty \)’ appropriately.
NOTE: The graph has a vertical asymptote \(x=-2\) and a horizontal asymptote \(y = 1\).
For Exercises factorcancelfirst - factorcancellast, find the limit analytically using Exercise rationallimit as a guide. If a limit fails to exist, state that is the case or use the
symbols ‘\(\infty \)’ or ‘\(-\infty \)’ appropriately.
For Exercises abscancelfirst - abscancellast, use the piecewise definition of absolute value, Definition absolutevaluepiecewise to help you find the limit analytically. If a limit fails
to exist, state that is the case or use the symbols ‘\(\infty \)’ or ‘\(-\infty \)’ appropriately.
For Exercises complexcancelfirst - complexcancellast, simplify the complex fraction in order to help you find the limit analytically. (A review of Section AppRatExpEqus may
be in order.) If a limit fails to exist, state that is the case or use the symbols ‘\(\infty \)’ or ‘\(-\infty \)’ appropriately.
For Exercises radicalcancelfirst - radicalcancellast, rationalize the numerator of the fraction in order to help you find the limit analytically. (A
review of Section rationalizingdenomandnumer may be in order.) If a limit fails to exist, state that is the case or use the symbols ‘\(\infty \)’ or ‘\(-\infty \)’
appropriately.
Let \(f(x) = \frac {x}{\lfloor x \rfloor }\) where ‘\(\lfloor x \rfloor \)’ is the greatest integer (or floor) function. (See Example greatestintegerdefn in Section ConstantFunctions for review, if needed.)
Fill in the blanks below to help you analyze \(\lim _{x \rightarrow 0} f(x)\).
If \(-1 < x < 0\), then \(\lfloor x \rfloor = \answer {-1}\). So we can rewrite \(f(x) = \frac {x}{\lfloor x \rfloor } = \answer {-x}\).
Using part (a), we can find \(\lim _{x \rightarrow 0^{-}} f(x) = \lim _{x \rightarrow 0^{-}} (\answer {-x}) = \answer {0}\).
If \(0 < x < 1\), then \(\lfloor x \rfloor =\answer {0}\), hence \(f(x) = \frac {x}{\lfloor x \rfloor }\) does not exist as \(x \rightarrow 0^{+}\).
Putting parts (b) and (c) together, we have that \(\lim _{x \rightarrow 0} f(x)\) does not exist.
Graph \(f(x) = \frac {x}{\lfloor x \rfloor }\) on desmos near \(x = 0\) to confirm your answers.
With help from your classmates, sketch the graph of a function which satisfies all of the following criteria:
\(\lim _{x \rightarrow -\infty } f(x) = \infty \)
\(\lim _{x \rightarrow 4^{-}} f(x) = 6\)
\(\lim _{x \rightarrow 4^{+}} f(x) = - \infty \)
\(\lim _{x \rightarrow \infty } f(x) =0\)
With help from your classmates, sketch the graph of a function \(f\) which satisfies all of the following criteria:
\(\lim _{x \rightarrow -\infty } f(x) = 2\)
\(\lim _{x \rightarrow 0^{-}} f(x) = \infty \)
\(\lim _{x \rightarrow 0^{+}} f(x) = -\infty \)
\(\lim _{x \rightarrow 2^{-}} f(x) = 3\)
\(\lim _{x \rightarrow 2^{+}} f(x) =0\)
\(\lim _{x \rightarrow \infty } f(x) = -\infty \)
A function is said to be continuous from the left at \(x=a\) if \(\lim _{x \rightarrow a^{-}} f(x) = f(a)\). Likewise, a function is said to be continuous from the right at \(x=a\) if
\(\lim _{x \rightarrow a^{+}} f(x) = f(a)\).
Explain why \(r(x) = \sqrt {5-x}\) is not continuous at \(x = 5\). Is \(r\) continuous from the left at \(x=5\)? From the right? Explain.
The function \(r(x) = \sqrt {5-x}\) is not continuous at \(x = 5\) since \(r\) is undefined if \(x>5\), so \(\lim _{x \rightarrow 5^{+}} r(x)\) does not exist. However, \(r\) is continuous from the left at \(x = 5\)
since
Explain why the floor function (See Example greatestintegerdefn in Section ConstantFunctions for review, if needed.)\(F(x) = \lfloor x \rfloor \) is not continuous at \(x = 117\). Is \(F\)
continuous from the left at \(x=117\)? From the right? Explain.
The function \(F(x) = \lfloor x \rfloor \) is not continuous at \(x = 117\) since \(\lim _{x \rightarrow 117^{-}} \lfloor x \rfloor = 116\) but \(\lim _{x \rightarrow 117^{+}} \lfloor x \rfloor = 117\). However, \(F\) is continuous from the right at \(x = 117\) since \(\lim _{x \rightarrow 117^{+}} \lfloor x \rfloor = 117 = \lfloor 117 \rfloor = F(117)\).
If a function \(f\) is continuous at \(x = a\), explain why \(f\) is continuous from both the left and the right at \(x = a\). Is the converse true? That
is, if \(f\) is continuous from both the left and the right at \(x = a\), is \(f\) continuous at \(x=a\)?
If \(f\) is continuous at \(x=a\), then \(\lim _{x \rightarrow a} f(x) = f(a)\). This means \(\lim _{x \rightarrow a^{-}} f(x) = f(a)\) and \(\lim _{x \rightarrow a^{+}} f(x) = f(a)\), so \(f\) is continuous from both directions at \(x=a\). The converse is also true since
if \(\lim _{x \rightarrow a^{-}} f(x) = f(a)\) and \(\lim _{x \rightarrow a^{+}} f(x) = f(a)\), then \(\lim _{x \rightarrow a} f(x) = f(a)\).
The difference between the scenario here and that in Exercise twosidedonesidedlimitexistexercise is that here, we know what each of the one-sided limits
are: \(f(a)\). They can’t be different numbers like they could be in Example limitfromgraphex.
It turns out that \(\lim _{x \rightarrow 0} f(x) = 117\). How is this possible assuming the data in the table is correct?
We are told nothing of the function on the interval \((-0.000001, 0.000001 )\) so the function has plenty of opportunities to approach \(117\).
In this exercise, we use Definition infinitelimitsatinfinity to prove \(\lim _{x \rightarrow \infty } x^3 = \infty \) and \(\lim _{x \rightarrow -\infty } x^3 = -\infty \):
Solve the following inequalities:
\(x^3 > 1000\)
\(x^3 > 1000\) for \(x > 10\)
\(x^3 > 100000\)
\(x^3 > 1000000\) for \(x > 100\)
\(x^3 > 10^{99}\)
\(x^3 > 10^{99}\) for \(x > 10^{33}\)
\(x^3 > N\)
\(x^3 > N\) for \(x > \sqrt [3]{N}\)
Show that for \(N > 0\), if \(x > \sqrt [3]{N}\), then \(x^3 > N\). Write a sentence (or two!) which uses your work and Definition infinitelimitsatinfinity to prove \(\lim _{x \rightarrow \infty } x^3 = \infty \).
If \(x > \sqrt [3]{N}\), then \(x^3 > \left (\sqrt [3]{N} \right )^3 = N\). Per Definition infinitelimitsatinfinity, given \(N>0\), choose \(M = \sqrt [3]{N}\). If \(x > M\), then \(x^3 > M^3 = N\). Hence, \(\lim _{x \rightarrow \infty } x^3 = \infty \).
Repeat a similar argument to prove \(\lim _{x \rightarrow -\infty } x^3 = -\infty \).
![[Picture]](ExercisesforIntroLimits.online-863e68663a81e4a66cf9b100eea3736d.svg)
![[Picture]](ExercisesforIntroLimits.online-ac03e12eaeb7b5bbb1bb43f7bce80ab1.svg)