Question

using the squeeze theorem, calculate this limit...

Answer 1

\[\lim_{n \rightarrow \infty} \sqrt[n]{\frac{ 6 }{ 2^{n} + 3^{n} }}\]

Answer 2

by just looking at this i can tell the limit is 1/7. (if i'm right). but i don't know how to use the squeeze theorem to calculate this. any help would be appreciated.

Answer 3

@phi

Answer 4

Really, 1/7?
What is the limit as n --> infty of \( \sqrt[n]{2^n + 3^n} \) ?

Answer 5

you can use your calculator to get a look as to what value of the limit will be, and you'd get 0.

Answer 6

here's the inequality that squeezes the expression

\[\frac{6}{3^n + 3^n}<\frac{6}{2^n+3^n}<\frac{6}{2^n+2^n}\]

Answer 7

Well, does that squeeze down to one number? I don't think so.

Answer 8

What is the limit as n --> infty of 2n+3n−−−−−−√n ? this is 5.

Answer 9

so i think the limit 1/5. sorry. just by looking at it. am i right? if so how do we do it with the squeeze theorem? @JamesJ

Answer 10

No, 3^n gets much much larger 2^n and dominates. You should expect that
\[ \lim_{n \rightarrow \infty} \sqrt[n]{2^n + 3^n} = 3 \]

Answer 11

Notice that
\[ (2^n + 3^n)^{1/n} = 3 \cdot ((2/3)^n + 1)^{1/n} \]
Can you see now how to squeeze this between terms that are or go to 3?

Answer 12

\[\frac{6}{2^n+2^n}=\frac{3}{2^n}\\\frac{\sqrt[n]{3}}{\sqrt[n]{2^n}}=\frac{\sqrt[n]{3}}{2}\]
as \(n\rightarrow \infty\), the expression approaches zero.

Answer 13

No, that's wrong. For all a > 0, a^(1/n) --> 1 as n --> infty

Answer 14

why did you pull 3 out? @JamesJ

Answer 15

To give you some insight and help you figure out how to write an expressions
\[ stuff \leq (2^n + 3^n)^{1/n} \leq stuff \]

Answer 16

you want now that stuff and stuff both tend to 3 as n --> infty

Answer 17

@JamesJ and what about the numerator when x approaches infinity|

Answer 18

so what happened to the numerator?

Answer 19

no first i want to know why you're ignoring the \[\sqrt[n]{6}\]

Answer 20

Because that's easy to deal with. That term goes to 1.

Answer 21

oh yeah, the limit is 1, not zero. my mistake. i was thinking of a geometric sequence. silly of me.

Answer 22

ok great. now. i think. \[\sqrt[n]{2^{n} + 3^{n}} = \sqrt[n]{2^{n}} + \sqrt[n]{3^{n}} = 2 + 3\] right or wrong? @JamesJ

Answer 23

Completely, horribly wrong.

Answer 24

:(. why?

Answer 25

That's like saying
13^(1/2) = (4 + 9)^(1/2) = 4^(1/2) + 9^(1/2) = 2 + 3 = 5
which is false.

Answer 26

3⋅((2/3)n+1)^1/n ok now i get this part. but i don't see how to squeeze it? :(. can you please help me? @JamesJ

Answer 27

What is the answer we expect again? What we want to squeeze it between?

Answer 28

to 3

Answer 29

Yes, so one obvious lower bound is 3 itself, because the complicated term in brackets is > 1. Hence
\[ 3 < 3 \cdot ((2/3)^n + 1)^{1/n} < stuff \]

You now think about what upper bound might work.

Answer 30

still going to be 3?

Answer 31

Whatever "stuff" is, it had better have limit of 3 as n --> infty

Answer 32

...but it clearly can NOT be just 3, because if it were, then 3 < 3 !!

Answer 33

this is where i'm stuck. cause i don't even know how to find lower/upper bound. :(

Answer 34

What is the limit of 2^(1/n) as n --> infinity ?

Answer 35

1

Answer 36

Yes, so if you can bound the stuff in brackets with 2^(1/n), you will have it

Answer 37

this might be 3x(2^(1/n) ??

Answer 38

? What you've written doesn't make sense. I don't know what you mean.

Answer 39

it doesn't make sense cause i don't understand it. that's why i'm here :(. i'm trying to understand it.

Answer 40

We are trying to finish this expression:
\[ 3 < 3 \cdot ((2/3)^n + 1)^{1/n} < stuff \]

I am suggesting to you that one way to write "stuff" is to involve a term like \( 2^{1/n} \). But there's more than one solution. Whatever else you have you want
\[ \lim_{n \rightarrow \infty} stuff = 3 \]

Now think hard and play around until you find something.

Answer 41

Hint: 1 + 1 = 2

Answer 42

Hint: (2/3)^n < 1

Answer 43

i can say that lim as n > inf of 3x2^(1/n) = 3

Answer 44

yes

Answer 45

so what do i do next? i put 3x2^(1/n) = "stuff"? (on the right side? )

Answer 46

yes

Answer 47

so final answer comes out to be 1/3?

Answer 48

yes

Answer 49

ok can you please help me on one more thing? how can i find questions involving limits (with square roots involved to practice. this type of questions keeps popping up in my tests and exams and i fail everytime. i can't find a place to learn it online :( @JamesJ

Answer 50

Look in your text.

And rework this problem with a blank piece of paper. When you can do that without looking back at your notes, you will learn something.

Answer 51

can you please help me solve this one? \[\sqrt[n]{2^{n} + 3^{n+1}}\] can i say the upper bound is \[\sqrt[n]{3^{n} + 3^{n+1}}\] @JamesJ

Answer 52

This is very very similar to the problem we just worked. What do you think the limit is?

Answer 53

Again, intuitively, 3^(n+1) will soon be massively big compared to 2^n, so intuitively you should expect that
\[ (2^n + 3^{n+1})^{1/n} \longrightarrow (3^{n+1})^{1/n} = 3 \cdot 3^{1/n} \rightarrow 3 \]

Answer 54

Now imitate the arguments we just painstakingly developed above to put lower and upper bounds on
\[ (2^n + 3^{n+1})^{1/n} \]

Answer 55

upper bound = \[\lim_{n \rightarrow \infty} \sqrt[n]{(3^{n+1})}\]

Answer 56

No, that's a lower bound, because
\[ 3^{n+1} < 2^n + 3^{n+1} \]
and hence
\[ \sqrt[n]{ 3^{n+1}} < \sqrt[n]{ 2^n + 3^{n+1}} \]

Answer 57

oops! mean to say lower bound.

Answer 58

now uper bound = \[\lim_{n \rightarrow \infty} 3 \times \sqrt[n]{2}\] @JamesJ

Answer 59

hm, not quite. You need to do some work and write it out a bit more.

Answer 60

\[\lim_{n \rightarrow \infty} \sqrt[n]{3^{n}} = 3\] how about that?

Answer 61

\[ (2^n + 3^{n+1})^{1/n} = 3 \cdot ( (2/3)^n + 3 )^{1/n} \]
Now see what to do?

Answer 62

no :(

Answer 63

You are trying to show that
\[ \lim_{n \rightarrow \infty} (2^n + 3^{n+1})^{1/n} = 3 \]
Just as with your first problem, the way to do this will be to bound this expression above and below to something easy which clearly have limits of 3.

The lower bound is straight forward:
\[ 3\cdot 3^{1/n} = 3^{(n+1)/n} < (2^n + 3^{n+1})^{1/n} \]
...and clearly
\[ \lim_{n \rightarrow \infty} 3\cdot 3^{1/n} = 3 \lim_{n \rightarrow \infty} \cdot 3^{1/n} = 3 \cdot 1 = 3 \]

Now what we just wrote above is the UPPER bound.

Answer 64

sorry, it's not the upper bound. But just like the first problem we had, there is a way to bound it above so that you can easily find the limit:
\[ 2/3 < 1 \Rightarrow (2/3)^n + 3 < 1 + 3 = 4 \]
Do you now see how to bound the expression (2^n + 3^n+1)^(1/n) above?

Answer 65

\[ (2^n + 3^{n+1})^{1/n} = 3 \cdot ( (2/3)^n + 3 )^{1/n} < 3 \cdot (1 + 3)^{1/n} = 3 \cdot 4^{1/n} \]
Got it now?

Answer 66

so that'll be the upper bound? 3X4^1/n?

Answer 67

yes

Answer 68

so. 2/3 < 1 so you just can REPLACE (2/3)^n with 1? this makes no sense to me.

Answer 69

We are not replacing (2/3)^n with 1, we are bounding (2/3)^n above by 1:
\[ 2/3 < 1 \Rightarrow (2/3)^n < 1^n =1 \]
hence
\[ (2/3)^n + 3 < 1 + 3 = 4 \]

Answer 70

ok. lookst like i have to understand what "bounding" means and HOW to do it. thanks for all the help. and sorry if i wasted your time.

Answer 71

Bounding above means finding a quantity which is greater, that's all. Bounding below means finding a quantity which is less.

Answer 72

i understand what you thought me. i just need to know HOW TO "BOUND" i don't get that concept.

Answer 73

As for *how* to do it, these sorts of things are idiosyncratic and takes experience. That's why you need to do exercises like this one.

Answer 74

but i noticed YOU find the limit BEFORE looking for the lower and upper bounds. why do you do that?

Answer 75

I figure it out intuitively so I know what we're aiming for. That helps me figure out what kinds of bounds will be useful.

Answer 76

so i can do the same thing in my tests and exams? look at for the limit FIRST...?

Answer 77

If I was writing out the solution formally, I wouldn't mention that intuitive argument first. That is just for me so I know what direction to head in. So yes: do figure out what you think the answer is, but no: don't write out that piece because it isn't formal mathematics.

Answer 78

yeap i get it. i'm looking for more examples like this to try. i'd be grateful if you can find me 2. i try to google "squeeze theorem problems" and all i get is problems involving trigs. which is very easy. and those don't come in my tests and exams. just these kind and i can't find them online...

Answer 79

Try finding the limit of this
\[ \sqrt[n]{3^{n + 17} + 5^{n}} \]

Answer 80

I'm going now. talk to you later.

Answer 81

ok answer will be ready by the time you come back. thanks a lot

Answer 82

and here's one more
\[ \sqrt[n/2]{\frac{7^{2n}}{8^{3n} + 9^n}} \]

Answer 83

first i want to say the limist of the first example is 8. do you agree?

Answer 84

@JamesJ

Answer 85

i found another solution
\[\frac{3}{3^n}=\frac{6}{2(3^n)}=\frac{6}{3^n+3^n}<\frac{6}{2^n+3^n}<\frac{6}{3^n}\]

\[\frac{\sqrt[n]{3}}{\sqrt[n]{3^n}}<\sqrt[n]{\frac{6}{2^n+3^n}}<\frac{\sqrt[n]{6}}{\sqrt[n]{3^n}}\]
\[\frac{\sqrt[n]{3}}{3}<\sqrt[n]{\frac{6}{2^n+3^n}}<\frac{\sqrt[n]{6}}{3}\]

\[\lim_{n\rightarrow \infty}\sqrt[n]{3}=\lim_{n\rightarrow \infty}\sqrt[n]{6} =1\]

Answer 86

you mean 1/3 to be the limit. @sirm3d

Answer 87

yes, just like jamesj gave.

Answer 88

With respect to this question, the limit is not 8.
\[ \lim_{n \rightarrow \infty} \sqrt[n]{3^{n + 17} + 5^{n}} \]

Maybe this data will help give you a sense of what is intuitively important here: the 5^n term quickly dominates:

Answer 89

This just like the case for
\[ \sqrt[n]{2^n + 3^n} \]
The 3^n term quickly dominates:

Answer 90

so the lower bound is 5?

Answer 91

Yes, 5 is a lower bound. We are driving towards the answer being 5, so 5 as a lower bound is super useful.

Answer 92

Write out the proof for why 5 is a lower bound.

Answer 93

Let's begin a new thread on this. It's hard to see. I'll post.

Answer 94

http://openstudy.com/study#/updates/511148fbe4b09cf125bdb137