Babynini
  • Babynini
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.
Mathematics
chestercat
  • chestercat
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Babynini
  • Babynini
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Babynini
  • Babynini
Babynini
  • Babynini

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zepdrix
  • zepdrix
Which piece is causing trouble? \(\large\rm \sqrt{x^3+x^2}\) or the \(\large\rm \sin\left(\frac{\pi}{x}\right)\)? :)
Babynini
  • Babynini
I'm not sure where to start xD
zepdrix
  • zepdrix
OpenStudy lag :O( lameeeee
Babynini
  • Babynini
hmm?
anonymous
  • anonymous
As we know, sin of any number will always be between -1 and 1, inclusive.
anonymous
  • anonymous
\[-1\le \sin \frac{ \pi }{ x }\le1\]
anonymous
  • anonymous
now we multiply whole thing by \[\sqrt{x ^{3}+x ^{2}}\]
Babynini
  • Babynini
multiply the whole sin(pi/x)?
zepdrix
  • zepdrix
Plug in x=0, which part of the function is causing a problem?
anonymous
  • anonymous
yes like this
anonymous
  • anonymous
its a squezz theorm @zepdrix
anonymous
  • anonymous
ok i ll do it step by step u ll understand let me finish
Babynini
  • Babynini
(@zepdrix the sin part?) okay @ayeshaafzal221 :)
zepdrix
  • zepdrix
..
anonymous
  • anonymous
\[-(\sqrt{x ^{3}+x ^{2}})\le \sqrt{x ^{3}+x ^{2}} \times \sin \frac{ \pi }{ x }\le \sqrt{x ^{3}+x ^{2}}\]
zepdrix
  • zepdrix
Yes, so start by putting bounds on the bad part. Sine function is bound by -1 and 1, yes? :)
Babynini
  • Babynini
Yeah
anonymous
  • anonymous
With an inequality like this, you can add "lim x->0" to all 3 parts and the expression will still be valid.
zepdrix
  • zepdrix
\[\large\rm -1\le \sin\left(\frac{\pi}{x}\right)\le 1\]
Babynini
  • Babynini
Okay, @zepdrix next what? :)
anonymous
  • anonymous
\[\lim_{x \rightarrow 0}-(\sqrt{x ^{3}+x ^{2}})\le \lim_{x \rightarrow 0} \sqrt{x ^{3}+x ^{2}} \times \sin \frac{ \pi }{ x }\le \lim_{x \rightarrow 0} \sqrt{x ^{3}+x ^{2}}\]
anonymous
  • anonymous
now you solve left hand side and then right hand side
zepdrix
  • zepdrix
:p
anonymous
  • anonymous
so left hand side is \[\lim_{x \rightarrow 0} -(\sqrt{0^{3} +0^{2}} =-\sqrt{0}=0\]
anonymous
  • anonymous
i am sub x=0
anonymous
  • anonymous
now right hand side \[\lim_{x \rightarrow 0}\sqrt{0^{3}+0^{2}}=0\]
anonymous
  • anonymous
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.
Babynini
  • Babynini
ooo so, to answer the question, to I have to write out that proof?
anonymous
  • anonymous
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
Babynini
  • Babynini
The question says to "use notation of the squeeze theorem"
anonymous
  • anonymous
yes :) thats what squeex theorm is
Babynini
  • Babynini
Okay, because I don't remember multiplying things in doing limits in class o.o
anonymous
  • anonymous
ur question clearly says use squeez theorm to show that expression =0
Babynini
  • Babynini
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
anonymous
  • anonymous
yup thats what i did
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
see these domain and range u must know by heart its really useful
anonymous
  • anonymous
in ur case just remember cos , sin and tan
Babynini
  • Babynini
ahh ok
Babynini
  • Babynini
I am beginning to understand.
anonymous
  • anonymous
good to know just tag if u have any other problem :)
Babynini
  • Babynini
ok thanks!!
Babynini
  • Babynini
i) -1
Babynini
  • Babynini
*sin(pi/x)
Babynini
  • Babynini
@ayeshaafzal221 does that all look correct?
anonymous
  • anonymous
yes @Babynini well done u got a hang of it :)
Babynini
  • Babynini
Thank you! :)

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