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Babynini
 one year ago
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.
Babynini
 one year ago
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.

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Babynini
 one year ago
Best ResponseYou've already chosen the best response.2dw:1443761963714:dw

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Which piece is causing trouble? \(\large\rm \sqrt{x^3+x^2}\) or the \(\large\rm \sin\left(\frac{\pi}{x}\right)\)? :)

Babynini
 one year ago
Best ResponseYou've already chosen the best response.2I'm not sure where to start xD

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0OpenStudy lag :O( lameeeee

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0As we know, sin of any number will always be between 1 and 1, inclusive.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[1\le \sin \frac{ \pi }{ x }\le1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now we multiply whole thing by \[\sqrt{x ^{3}+x ^{2}}\]

Babynini
 one year ago
Best ResponseYou've already chosen the best response.2multiply the whole sin(pi/x)?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Plug in x=0, which part of the function is causing a problem?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its a squezz theorm @zepdrix

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok i ll do it step by step u ll understand let me finish

Babynini
 one year ago
Best ResponseYou've already chosen the best response.2(@zepdrix the sin part?) okay @ayeshaafzal221 :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[(\sqrt{x ^{3}+x ^{2}})\le \sqrt{x ^{3}+x ^{2}} \times \sin \frac{ \pi }{ x }\le \sqrt{x ^{3}+x ^{2}}\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Yes, so start by putting bounds on the bad part. Sine function is bound by 1 and 1, yes? :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0With an inequality like this, you can add "lim x>0" to all 3 parts and the expression will still be valid.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0\[\large\rm 1\le \sin\left(\frac{\pi}{x}\right)\le 1\]

Babynini
 one year ago
Best ResponseYou've already chosen the best response.2Okay, @zepdrix next what? :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow 0}(\sqrt{x ^{3}+x ^{2}})\le \lim_{x \rightarrow 0} \sqrt{x ^{3}+x ^{2}} \times \sin \frac{ \pi }{ x }\le \lim_{x \rightarrow 0} \sqrt{x ^{3}+x ^{2}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now you solve left hand side and then right hand side

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so left hand side is \[\lim_{x \rightarrow 0} (\sqrt{0^{3} +0^{2}} =\sqrt{0}=0\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now right hand side \[\lim_{x \rightarrow 0}\sqrt{0^{3}+0^{2}}=0\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thus 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.

Babynini
 one year ago
Best ResponseYou've already chosen the best response.2ooo so, to answer the question, to I have to write out that proof?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if 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

Babynini
 one year ago
Best ResponseYou've already chosen the best response.2The question says to "use notation of the squeeze theorem"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes :) thats what squeex theorm is

Babynini
 one year ago
Best ResponseYou've already chosen the best response.2Okay, because I don't remember multiplying things in doing limits in class o.o

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ur question clearly says use squeez theorm to show that expression =0

Babynini
 one year ago
Best ResponseYou've already chosen the best response.2Yeah. 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yup thats what i did

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0see these domain and range u must know by heart its really useful

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0in ur case just remember cos , sin and tan

Babynini
 one year ago
Best ResponseYou've already chosen the best response.2I am beginning to understand.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0good to know just tag if u have any other problem :)

Babynini
 one year ago
Best ResponseYou've already chosen the best response.2i) 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)

Babynini
 one year ago
Best ResponseYou've already chosen the best response.2@ayeshaafzal221 does that all look correct?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes @Babynini well done u got a hang of it :)
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