1.5 Limits Of Trig Functionsap Calculus

The Six Basic Trigonometric Functions

Trigonometric functions allow us to use angle measures, in radians or degrees, to find the coordinates of a point on any circle—not only on a unit circle—or to find an angle given a point on a circle. They also define the relationship among the sides and angles of a triangle.

To define the trigonometric functions, first consider the unit circle centered at the origin and a point (P=(x,y)) on the unit circle. Let (θ) be an angle with an initial side that lies along the positive (x)-axis and with a terminal side that is the line segment (OP). An angle in this position is said to be in standard position (Figure (PageIndex{2})). We can then define the values of the six trigonometric functions for (θ) in terms of the coordinates (x) and (y).

Calculus 10th Edition answers to Chapter 1 - Limits and Their Properties - Section Project - Graphs and Limits of Trigonometric Functions - Page 90 c including work step by step written by community members like you. Textbook Authors: Larson, Ron; Edwards, Bruce H., ISBN-10: 1-28505-709-0, ISBN-13: 978-1-28505-709-5, Publisher: Brooks Cole. Hello, welcome back to www.educator.com and welcome back to AP Calculus.0000. Today, we are going to be talking about the derivatives of the trigonometric functions.0004. Let us jump right on in.0008. Dmg supplement vs tmg manufacturing. Let us start off by finding the derivative of sin x.0010. Let me see, I will stick with blue today.0016. Let us find the derivative of f(x) = the.

Definition: Trigonometric functions

Let (P=(x,y)) be a point on the unit circle centered at the origin (O). Let (θ) be an angle with an initial side along the positive (x)-axis and a terminal side given by the line segment (OP). The trigonometric functions are then defined as

If (x=0, secθ) and (tan θ) are undefined. If (y=0), then (cot θ) and (cscθ) are undefined.

We can see that for a point (P=(x,y)) on a circle of radius (r) with a corresponding angle (θ), the coordinates (x) and (y) satisfy

[begin{align} cos θ &=dfrac{x}{r} x&=rcos θ end{align}]

and

[begin{align} sin θ&=dfrac{y}{r} y&=rsin θ. end{align}]

Figure (PageIndex{3.1}): Diagram demonstrating trigonometric functions in the unit circle., ).

The values of the other trigonometric functions can be expressed in terms of (x,y), and (r) (Figure (PageIndex{3})).

Figure (PageIndex{3.2}): For a point (P=(x,y)) on a circle of radius (r), the coordinates (x) and y satisfy (x=rcos θ) and (y=rsin θ).

Table shows the values of sine and cosine at the major angles in the first quadrant. From this table, we can determine the values of sine and cosine at the corresponding angles in the other quadrants. The values of the other trigonometric functions are calculated easily from the values of (sin θ) and (cos θ.)

Example (PageIndex{2}): Evaluating Trigonometric Functions

Evaluate each of the following expressions.

1. (sin(dfrac{2π}{3}))
2. (cos(−dfrac{5π}{6}))
3. (tan(dfrac{15π}{4}))

Solution:

a) On the unit circle, the angle (θ=dfrac{2π}{3}) corresponds to the point ((−dfrac{1}{2},dfrac{sqrt{3}}{2})). Therefore, (sin)((dfrac{2π}{3}))=(y)=(dfrac{sqrt{3}}{2}).

b) An angle (θ=−dfrac{5π}{6}) corresponds to a revolution in the negative direction, as shown. Therefore, (cos(−dfrac{5π}{6}))=(x)=(−dfrac{sqrt{3}}{2}).

c) An angle (θ)=(dfrac{15π}{4})=(2π)+(dfrac{7π}{4}). Therefore, this angle corresponds to more than one revolution, as shown. Knowing the fact that an angle of (dfrac{7π}{4}) corresponds to the point ((dfrac{sqrt{2}}{2},dfrac{sqrt{2}}{2})), we can conclude that (tan(dfrac{15π}{4})=dfrac{y}{x}=−1).

Exercise (PageIndex{2})

Evaluate (cos(3π/4)) and (sin(−π/6)).

Look at angles on the unit circle.

Answer

While the idea of a sequence of numbers, $a_1,a_2,a_3,ldots$ isstraightforward, it is useful to think of a sequence as a function. Wehave up until now dealt with functions whose domains are the realnumbers, or a subset of the real numbers, like $f(x)=sin x$. Asequence is a function with domain the natural numbers$N={1,2,3,ldots}$ or the non-negative integers,$ds Z^{ge0}={0,1,2,3,ldots}$. The range of the function is stillallowed to be the real numbers; in symbols, we say that a sequence isa function $fcolon NtoR$. Sequences are written in a few differentways, all equivalent; these all mean the same thing:$$ displaylines{a_1,a_2,a_3,ldotscr left{a_nright}_{n=1}^inftycr left{f(n)right}_{n=1}^inftycr}$$

As with functions on the real numbers,we will most often encounter sequences that can be expressed by aformula. We have already seen the sequence $ds a_i=f(i)=1-1/2^i$, and othersare easy to come by:$$eqalign{ f(i)&={iover i+1}cr f(n)&={1over2^n}cr f(n)&=sin(npi/6)cr f(i)&={(i-1)(i+2)over2^i}cr}$$Frequently these formulas will make sense if thought of either asfunctions with domain $R$ or $N$, though occasionally one will makesense only for integer values.

Faced with a sequence we are interested in the limit$$lim_{ito infty} f(i) = lim_{itoinfty} a_i.$$We already understand$$lim_{xtoinfty} f(x)$$when $x$ is a real valued variable; now we simply want to restrict the'input' values to be integers. No real difference is required in thedefinition of limit, except that we specify, perhaps implicitly, thatthe variable is an integer. Compare this definition todefinition 4.10.4.

Definition 11.1.1 Suppose that $dsleft{a_nright}_{n=1}^infty$ is a sequence.We say that $ds lim_{nto infty}a_n=L$ if for every $epsilon>0$there is an $N > 0$ so that whenever $n>N$, $|a_n-L|< epsilon$. If$ds lim_{ntoinfty}a_n=L$ we say that the sequence converges,otherwise it diverges.

1.5 Limits Of Trig Functions Ap Calculus Calculator

If $f(i)$ defines a sequence, and $f(x)$ makes sense, and $ds lim_{xtoinfty}f(x)=L$, then it is clear that$ds lim_{itoinfty}f(i)=L$ as well, but it is important to note thatthe converse of this statement is not true. For example, since$ds lim_{xtoinfty}(1/x)=0$, it is clear that also$ds lim_{itoinfty}(1/i)=0$, that is, the numbers$${1over1},{1over2},{1over3},{1over4},{1over5},{1over6},ldots$$get closer and closer to 0. Consider this, however: Let $f(n)=sin(npi)$. This is the sequence$$ sin(0pi), sin(1pi),sin(2pi),sin(3pi),ldots=0,0,0,0,ldots$$since $sin(npi)=0$ when $n$ is an integer. Thus$ds lim_{ntoinfty}f(n)=0$. But $ds lim_{xtoinfty}f(x)$, when $x$ isreal, does not exist: as $x$ gets bigger and bigger, the values$sin(xpi)$ do not get closer and closer to a single value, but takeon all values between $-1$ and $1$ over and over. In general, wheneveryou want to know $ds lim_{ntoinfty}f(n)$ you should first attempt tocompute $ds lim_{xtoinfty}f(x)$, since if the latter exists it is alsoequal to the first limit. But if for some reason$ds lim_{xtoinfty}f(x)$ does not exist, it may still be true that $ds lim_{ntoinfty}f(n)$ exists, but you'll have to figure outanother way to compute it.

It is occasionally useful to think of the graph of a sequence. Sincethe function is defined only for integer values, the graph is just asequence of dots. In figure 11.1.1 we see thegraphs of two sequences and the graphs of the corresponding realfunctions.

Not surprisingly, the properties of limits of real functions translateinto properties of sequences quite easily. Theorem 2.3.6 about limits becomes

Theorem 11.1.2 Suppose that $dslim_{ntoinfty}a_n=L$ and $dslim_{ntoinfty}b_n=M$ and$k$ is some constant. Then$$eqalign{&lim_{ntoinfty} ka_n = klim_{ntoinfty}a_n=kLcr&lim_{ntoinfty} (a_n+b_n) = lim_{ntoinfty}a_n+lim_{ntoinfty}b_n=L+Mcr&lim_{ntoinfty} (a_n-b_n) = lim_{ntoinfty}a_n-lim_{ntoinfty}b_n=L-Mcr&lim_{ntoinfty} (a_nb_n) = lim_{ntoinfty}a_ncdotlim_{ntoinfty}b_n=LMcr&lim_{ntoinfty} {a_nover b_n} = {lim_{ntoinfty}a_nover lim_{ntoinfty}b_n}={Lover M},hbox{ if $M$ is not 0}cr}$$

Likewise the Squeeze Theorem (4.3.1) becomes

Theorem 11.1.3 Suppose that $ds a_n le b_n le c_n$ for all $n>N$, for some $N$.If $dslim_{ntoinfty}a_n=dslim_{ntoinfty}c_n=L$, then $dslim_{ntoinfty}b_n=L$.

1.5 Limits Of Trig Functionsap Calculus Identities

And a final useful fact:

Theorem 11.1.4 $dslim_{ntoinfty}|a_n|=0$ if and only if$dslim_{ntoinfty}a_n=0$.

This says simply that the size of $ds a_n$ gets close to zero if andonly if $ds a_n$ gets close to zero.

Example 11.1.5 Determine whether $dsleft{{nover n+1}right}_{n=0}^infty$ converges ordiverges. If it converges, compute the limit. Since this makes sensefor real numbers we consider$$lim_{xtoinfty}{xover x+1}=lim_{xtoinfty}1-{1over x+1}=1-0=1.$$Thus the sequence converges to 1.

Example 11.1.6 Determine whether $dsbigg{{ln nover n}bigg}_{n=1}^infty$ converges ordiverges. If it converges, compute the limit. We compute$$lim_{xtoinfty}{ln xover x}=lim_{xtoinfty}{1/xover 1}=0,$$using L'Hôpital's Rule. Thus the sequence converges to 0.

Example 11.1.7 Determine whether $ds{(-1)^n}_{n=0}^infty$ converges ordiverges. If it converges, compute the limit. This does not make sensefor all real exponents, but the sequence is easy to understand: it is$$1,-1,1,-1,1ldots$$and clearly diverges.

Example 11.1.8 Determine whether $ds{(-1/2)^n}_{n=0}^infty$ converges ordiverges. If it converges, compute the limit. We consider the sequence $ds{|(-1/2)^n|}_{n=0}^infty={(1/2)^n}_{n=0}^infty$.Then$$ lim_{xtoinfty}left({1over2}right)^x=lim_{xtoinfty}{1over2^x}=0,$$so by theorem 11.1.4 the sequence converges to 0.

Example 11.1.9 Determine whether $ds{(sin n)/sqrt{n}}_{n=1}^infty$ converges ordiverges. If it converges, compute the limit. Since $|sin n|le 1$, $ds 0le|sin n/sqrt{n}|le1/sqrt{n}$ and we can use theorem 11.1.3 with $ds a_n=0$ and $ds c_n=1/sqrt{n}$. Since$dslim_{ntoinfty} a_n=dslim_{ntoinfty} c_n=0$, $dslim_{ntoinfty}sin n/sqrt{n}=0$ and the sequence converges to 0.

Example 11.1.10 A particularly common and useful sequence is $ds {r^n}_{n=0}^infty$,for various values of $r$. Some are quite easy to understand: If $r=1$the sequence converges to 1 since every term is 1, and likewise if$r=0$ the sequence converges to 0. If $r=-1$ this isthe sequence of example 11.1.7 anddiverges. If $r>1$ or $r< -1$ the terms $ds r^n$ get large without limit,so the sequence diverges. If $0< r< 1$ then the sequence converges to0. If $-1< r< 0$ then $ds |r^n|=|r|^n$ and $0< |r|< 1$, so the sequence$ds {|r|^n}_{n=0}^infty$ converges to 0, so also $ds{r^n}_{n=0}^infty$ converges to 0.converges. In summary, $ds {r^n}$ converges precisely when$-1< rle1$ in which case$$ lim_{ntoinfty} r^n=cases{ 0& if $-1< r< 1$cr 1& if $r=1$cr}$$

Sometimes we will not be able to determine the limit of a sequence,but we still would like to know whether it converges. In some cases wecan determine this even without being able to compute the limit.

A sequence is called increasingor sometimes strictly increasing if $ds a_i< a_{i+1}$ for all$i$. It is called non-decreasing or sometimes(unfortunately) increasing if $ds a_ile a_{i+1}$ for all$i$. Similarly a sequence is decreasing if $ds a_i>a_{i+1}$ for all $i$and non-increasing if$ds a_ige a_{i+1}$ for all $i$. If a sequence has any of theseproperties it is called monotonic.

Example 11.1.11 The sequence$$ left{{2^i-1over2^i}right}_{i=1}^infty= {1over2},{3over4},{7over8},{15over16},ldots,$$is increasing, and$$ left{{n+1over n}right}_{i=1}^infty= {2over1},{3over2},{4over3},{5over4},ldots$$is decreasing.

1.5 Limits Of Trig Functions Ap Calculus Integrals

A sequence is bounded aboveif there is some number $N$ such that $ds a_nle N$ for every $n$,and bounded below if there issome number $N$ such that $ds a_nge N$ for every $n$. If a sequenceis bounded above and bounded below it is bounded. If a sequence $ds{a_n}_{n=0}^infty$ is increasing or non-decreasing it is boundedbelow (by $ds a_0$), and if it is decreasing or non-increasing it isbounded above (by $ds a_0$). Finally, with all this new terminologywe can state an important theorem.

Theorem 11.1.12 If a sequence is bounded and monotonic then it converges.

We will not prove this; the proof appears in many calculus books. Itis not hard to believe: suppose that a sequence is increasing andbounded, so each term is larger than the one before, yet never largerthan some fixed value $N$. The terms must then get closer and closerto some value between $ds a_0$ and $N$. It need not be $N$, since $N$ maybe a 'too-generous' upper bound; the limit will be thesmallest number that is above all of the terms $ds a_i$.

Example 11.1.13 All of the terms $ds (2^i-1)/2^i$ are less than 2, and the sequence isincreasing. As we have seen, the limit of the sequence is 1—1 is thesmallest number that is bigger than all the terms in the sequence.Similarly, all of the terms $(n+1)/n$ are bigger than $1/2$, and thelimit is 1—1 is the largest number that is smaller than the terms ofthe sequence.

We don't actually need to know that a sequence is monotonic to applythis theorem—it is enough to know that the sequence is'eventually' monotonic, that is, that at some point it becomesincreasing or decreasing. For example, the sequence $10$, $9$, $8$,$15$, $3$, $21$, $4$, $3/4$, $7/8$, $15/16$, $31/32,ldots$ is notincreasing, because among the first few terms it is not. But startingwith the term $3/4$ it is increasing, so the theorem tells us that thesequence $3/4, 7/8, 15/16, 31/32,ldots$ converges. Since convergencedepends only on what happens as $n$ gets large, adding a fewterms at the beginning can't turn a convergent sequence into adivergent one.

Example 11.1.14 Show that $ds{n^{1/n}}$ converges.

1.5 Limits Of Trig Functions Ap Calculus Algebra

We first show that this sequence is decreasing, that is, that $ds n^{1/n}>(n+1)^{1/(n+1)}$. Consider the real function $ds f(x)=x^{1/x}$ when$xge1$. We can compute the derivative, $ds f'(x)=x^{1/x}(1-ln x)/x^2$,and note that when $xge 3$ this is negative. Since the function hasnegative slope, $ds n^{1/n}>(n+1)^{1/(n+1)}$ when $nge 3$. Since all terms of the sequence arepositive, the sequence is decreasing and bounded when $nge3$, and sothe sequence converges. (As it happens, we can compute the limit inthis case, but we know it converges even without knowing the limit; seeexercise 1.)

Example 11.1.15 Show that $ds{n!/n^n}$ converges.

Again we show that the sequence is decreasing, and since each term ispositive the sequence converges. We can't take the derivative thistime, as $x!$ doesn't make sense for $x$ real. But we note that if $ds a_{n+1}/a_n < 1$ then $ds a_{n+1}< a_n$, which is what we want toknow. So we look at $ds a_{n+1}/a_n$:$$ {a_{n+1}over a_n} = {(n+1)!over (n+1)^{n+1}}{n^nover n!}= {(n+1)!over n!}{n^nover (n+1)^{n+1}}= {n+1over n+1}left({nover n+1}right)^n= left({nover n+1}right)^n < 1.$$(Again it is possible to compute the limit; seeexercise 2.)

Exercises 11.1

1.5 Limits Of Trig Functions Ap Calculus Multiple Choice

Ex 11.1.1Compute $dslim_{xtoinfty} x^{1/x}$.(answer)

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Ex 11.1.2Use the squeeze theorem to show that $dslim_{ntoinfty} {n!over n^n}=0$.

Ex 11.1.3Determine whether $ds{sqrt{n+47}-sqrt{n}}_{n=0}^infty$ converges or diverges. If it converges, compute the limit.(answer)

Ex 11.1.4Determine whether $dsleft{{n^2+1over (n+1)^2}right}_{n=0}^infty$ converges or diverges. If it converges, compute the limit.(answer)

Ex 11.1.5Determine whether $dsleft{{n+47oversqrt{n^2+3n}}right}_{n=1}^infty$ converges or diverges. If it converges, compute the limit.(answer)

Ex 11.1.6Determine whether $dsleft{{2^nover n!}right}_{n=0}^infty$ converges or diverges. (answer)