# 1.5 Limits Of Trig Functionsap Calculus

Dec 21, 2020 1.0: Library of functions. 1.0E: Exercises; 1.1: Introduction to concept of a limit. 1.1E: Exercises; 1.2: One sided Limits and Vertical Asymptotes. 1.5 Limits of Trig Functions. Powered by Create your own unique website with customizable templates. Free Calculus worksheets created with Infinite Calculus. Printable in convenient PDF format. Math AP®︎/College Calculus AB Limits and continuity Determining limits using algebraic properties of limits: direct substitution Limits of trigonometric functions AP.CALC: LIM‑1 (EU), LIM‑1.D (LO), LIM‑1.D.1 (EK). Limits of integral are from θ = π to θ = 0. Reversing the limits changes the minus back to plus: 1 dx π −1 1 − x 1 2 √ 1 − x2 = 0 π sin θdθ dx √ 1 − x2 = 0 dθ = π −1 (The substitution x = sin t works similarly, but the limits of integration are −π/2 and π/2.) c) (x = sin t, dx = cos tdt) 1 1 1 π/2 π/2.

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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**.

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 $dslim_{ntoinfty}b_n=L$.

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.

## 1.5 Limits Of Trig Functions Ap Calculus Calculator

A sequence is called **increasing**or 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**.

## 1.5 Limits Of Trig Functions Ap Calculus Algebra

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.

A sequence is **bounded above**if 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.

## Limits Of Trig Functions Pdf

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.

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. American gangster french.

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.)

## 1.5 Limits Of Trig Functions Ap Calculus Frq

## Exercises 11.1

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

## 1.5 Limits Of Trig Functionsap Calculus Solver

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

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

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

## Limits Of Trig Functions Infinity

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

## 1.5 Limits Of Trig Functionsap Calculus Identities

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