anonymous
  • anonymous
evaluate this limit
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\lim_{x \rightarrow 5}(\frac{ 1 }{ x }-\frac{ 1 }{ 5 })(\frac{ 1 }{ x^3-4x^2-5x })\]
anonymous
  • anonymous
0?
anonymous
  • anonymous
seems why you graph it it tends to -1/750

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anonymous
  • anonymous
when*
terenzreignz
  • terenzreignz
I wonder... could we instead, reduce it to a single fraction bar? So much simpler that way...
anonymous
  • anonymous
(5-x)/(5x*(x^3 - 4x^2 -5x)), i think it is infinity cuz the denominator is zero too when x->5
anonymous
  • anonymous
\[\lim_{x \rightarrow 5}\frac{ 5-x }{ 5x(x^3-4x^2-5x) }\]
terenzreignz
  • terenzreignz
\[\lim_{x \rightarrow 5}\left(\frac{5-x}{5x}\right)\left(\frac{ 1 }{ x^3-4x^2-5x }\right)\] \[\lim_{x \rightarrow 5}\left(\frac{5-x}{5x( x^3-4x^2-5x)}\right)\] \[\Large\lim_{x \rightarrow 5}\left(\frac{5-x}{5 x^4-20x^3-25x^2}\right)\]
anonymous
  • anonymous
seperate now?
terenzreignz
  • terenzreignz
Are we, perhaps, allowed to use L'Hôpital?
anonymous
  • anonymous
nup!
anonymous
  • anonymous
says not to
terenzreignz
  • terenzreignz
ohh.. Well, no problem :D
anonymous
  • anonymous
wait it doesn't say u need to or not. thats for the other question
terenzreignz
  • terenzreignz
We just have to identify the largest exponent which x is given, in this case, 4 \[\Large\lim_{x \rightarrow 5}\left(\frac{5-x}{5\color{red}{ x^4}-20x^3-25x^2}\right)\] right?
anonymous
  • anonymous
now u divide everything by x^4?
terenzreignz
  • terenzreignz
Let's be formal... we multiply everything by... \[\LARGE \frac{\frac1{x^4}}{\frac1{x^4}}\]
anonymous
  • anonymous
oh k fair.
anonymous
  • anonymous
i get 0?
terenzreignz
  • terenzreignz
On second thought, that may not work out as intended... recalculating :3
terenzreignz
  • terenzreignz
I forgot it wasn't a limit to infinity... sorry, old habits die hard :D
anonymous
  • anonymous
haha
terenzreignz
  • terenzreignz
Start from here... \[\LARGE \lim_{x \rightarrow 5}\quad\frac{ 5-x }{ 5x(x^3-4x^2-5x) }\]
terenzreignz
  • terenzreignz
And factor the denominator.
terenzreignz
  • terenzreignz
If they are both zero at x = 5, then that must mean x-5 is a factor of the denominator.
anonymous
  • anonymous
\[\lim_{x \rightarrow 5}\frac{ 5-x }{ 5x(x-5)(x)(x+1) }\]
terenzreignz
  • terenzreignz
I bet you can see where this is going now? :)
anonymous
  • anonymous
\[\lim_{x \rightarrow 5}\frac{ 5-x }{ 5x^2(x-5)(x+1) }\]
anonymous
  • anonymous
YEP
anonymous
  • anonymous
i can just put a negaitve and change it to x-5
terenzreignz
  • terenzreignz
Yeah, sorry for that false lead :) Let me just make that tweak, to, err, make amends XD \[\lim_{x \rightarrow 5}\frac{ -1(x-5) }{ 5x^2(x-5)(x+1) }\]
anonymous
  • anonymous
haha thats why graphing is good! yep it comes to -1/750
anonymous
  • anonymous
gooood!
anonymous
  • anonymous
what about this one!
anonymous
  • anonymous
\[\lim_{x \rightarrow 0}\frac{ \tan4x }{ 5x^2+2x }\]
anonymous
  • anonymous
no L'Hopital though
terenzreignz
  • terenzreignz
Ooooh... let's have a look...
anonymous
  • anonymous
we can factor out x?
anonymous
  • anonymous
then we have sin4x/x i think i know where this might go to!
terenzreignz
  • terenzreignz
okay... tell me...
anonymous
  • anonymous
gimmi a sec
anonymous
  • anonymous
\[\lim_{x \rightarrow 0}\frac{ \sin4x }{ x }\frac{ 1 }{ \cos4x(5x+2) }\]
anonymous
  • anonymous
now we multiply both numerator and denominator by 4
terenzreignz
  • terenzreignz
Just go on...
terenzreignz
  • terenzreignz
While I silently ponder whether or not you needed my help with this question at all XD
anonymous
  • anonymous
\[\lim_{x \rightarrow 0}\frac{ \sin4x }{ 4x }\frac{ 4 }{ \cos4x(5x+2) }\] \[\lim_{4x \rightarrow 0}\frac{ \sin4x }{ 4x }\lim_{x \rightarrow 0}\frac{ 4 }{ \cos4x(5x+2) }\]
terenzreignz
  • terenzreignz
Of course, that second line is only possible because both limits exist ;)
terenzreignz
  • terenzreignz
And so the limit of the entire thingy is...?
anonymous
  • anonymous
2!
anonymous
  • anonymous
ah yes! since both limits exist. thats important
terenzreignz
  • terenzreignz
And... you're right. So... remind me again what you needed me for? ^_^
anonymous
  • anonymous
hahaha re-assurance!
anonymous
  • anonymous
and ur crazy good!
anonymous
  • anonymous
except ucouldn't help me with that linear algebra proof :( i think i nutted it out though not sure
terenzreignz
  • terenzreignz
I don't really do linear algebra... matrices hate me ^_^
anonymous
  • anonymous
haha ohh well! thanks for that! terenz to the rescue once again!
terenzreignz
  • terenzreignz
^_^

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