Question

Use squeeze theorem to show that
lim sqroot (x^3+x^2)sin(pi/x)=0
x-> 0
illustrate by graphing the functions f,g,h on the same screen.

Answer 1

@zepdrix

Answer 2

@Astrophysics

Answer 3

Which piece is causing trouble?
\(\large\rm \sqrt{x^3+x^2}\) or the \(\large\rm \sin\left(\frac{\pi}{x}\right)\)?
:)

Answer 4

I'm not sure where to start xD

Answer 5

OpenStudy lag :O( lameeeee

Answer 6

hmm?

Answer 7

As we know, sin of any number will always be between -1 and 1, inclusive.

Answer 8

\[-1\le \sin \frac{ \pi }{ x }\le1\]

Answer 9

now we multiply whole thing by \[\sqrt{x ^{3}+x ^{2}}\]

Answer 10

multiply the whole sin(pi/x)?

Answer 11

Plug in x=0, which part of the function is causing a problem?

Answer 12

yes like this

Answer 13

its a squezz theorm @zepdrix

Answer 14

ok i ll do it step by step u ll understand let me finish

Answer 15

(@zepdrix the sin part?)
okay @ayeshaafzal221 :)

Answer 16

..

Answer 17

\[-(\sqrt{x ^{3}+x ^{2}})\le \sqrt{x ^{3}+x ^{2}} \times \sin \frac{ \pi }{ x }\le \sqrt{x ^{3}+x ^{2}}\]

Answer 18

Yes, so start by putting bounds on the bad part.
Sine function is bound by -1 and 1, yes? :)

Answer 19

Yeah

Answer 20

With an inequality like this, you can add "lim x->0" to all 3 parts and the expression will still be valid.

Answer 21

\[\large\rm -1\le \sin\left(\frac{\pi}{x}\right)\le 1\]

Answer 22

Okay, @zepdrix next what? :)

Answer 23

\[\lim_{x \rightarrow 0}-(\sqrt{x ^{3}+x ^{2}})\le \lim_{x \rightarrow 0} \sqrt{x ^{3}+x ^{2}} \times \sin \frac{ \pi }{ x }\le \lim_{x \rightarrow 0} \sqrt{x ^{3}+x ^{2}}\]

Answer 24

now you solve left hand side and then right hand side

Answer 25

:p

Answer 26

so left hand side is \[\lim_{x \rightarrow 0} -(\sqrt{0^{3} +0^{2}} =-\sqrt{0}=0\]

Answer 27

i am sub x=0

Answer 28

now right hand side \[\lim_{x \rightarrow 0}\sqrt{0^{3}+0^{2}}=0\]

Answer 29

thus the limit of the original expression is less than and equal to 0, and greater and equal to 0, then its limit must be 0.

Answer 30

ooo so, to answer the question, to I have to write out that proof?

Answer 31

if u didnt understood any step let me know and i know what you confuse about whenever there is squeez theorm there are restriction that u must know before hand for cos , tan and sin , u can google em

Answer 32

The question says to "use notation of the squeeze theorem"

Answer 33

yes :) thats what squeex theorm is

Answer 34

Okay, because I don't remember multiplying things in doing limits in class o.o

Answer 35

ur question clearly says use squeez theorm to show that expression =0

Answer 36

Yeah. Usually we approach the limit from the left and right and see if they match up o.0 or calculate it as it approaches 0

Answer 37

yup thats what i did

Answer 38

see these domain and range u must know by heart its really useful

Answer 39

in ur case just remember cos , sin and tan

Answer 40

ahh ok

Answer 41

I am beginning to understand.

Answer 42

good to know just tag if u have any other problem :)

Answer 43

ok thanks!!

Answer 44

i) -1 </= sin (pi/x) </= 1
ii) \[-\sqrt{x^3+x^2}\le \sqrt{x^3+x^2}\sin(\frac{ \pi }{ x })\le \sqrt{x^3+x^2}\]
iii) we know that \[\lim \sqrt{x^3+^2} = 0 \space and \space \lim(-\sqrt{x^3+x^2}=0\]

Taking f(x)= -sq( x^3+x^2), g(x)= sq(x^3+x^2)sin(pi/x) and, h(x) = sq(x^3+x^2), in squeeze theorem, we conclude:

lim sq(x^3+x^2)sin(pi/)=0
(as approaches 0)

Answer 45

*sin(pi/x)

Answer 46

@ayeshaafzal221 does that all look correct?

Answer 47

yes @Babynini well done u got a hang of it :)

Answer 48

Thank you! :)