Zenmo
  • Zenmo
Calculate limits (Question on indeterminate forms).
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Zenmo
  • Zenmo
\[\lim_{x \rightarrow 0^+}(1+\sin4x)^{cotx}\] How is 1+sin4x --> 1 and cotx --> infinity? In order to know if the given limit is indeterminate.
SolomonZelman
  • SolomonZelman
No I am just trying to find the way, I don't know how to do this limit at this point asides from numerical approach.
Zenmo
  • Zenmo
Yea, I know how to solve it. But, I have to check if its indeterminate or not. Thanks for trying though. :)

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SolomonZelman
  • SolomonZelman
that indeterminate form doesn't seem to do much to me the way the limit is now...
Zenmo
  • Zenmo
1 Attachment
Zenmo
  • Zenmo
I'm reading a book, trying to make a sense out of it.
SolomonZelman
  • SolomonZelman
But 1+sin(4x) is not 1
SolomonZelman
  • SolomonZelman
That is 0 to 2, I checked via wolfr
SolomonZelman
  • SolomonZelman
so 2^∞=∞, 0^∞=0
zepdrix
  • zepdrix
? 0_o At x=0: 1+sin(4x) = 1, ya? :o
SolomonZelman
  • SolomonZelman
\(\large\color{#003366}{\displaystyle y=\lim_{x \rightarrow ~\infty }\left(1+\sin 4x\right)^{\cot(x)} }\) ln both sides, tnx for the link :)
Zenmo
  • Zenmo
ok so x=0: 1+sin(4x)=1. I got that, but \[\cot(x)=\]?
Zenmo
  • Zenmo
cot(x)=positive infinity*
SolomonZelman
  • SolomonZelman
\(\large\color{#003366}{\displaystyle y=\lim_{x \rightarrow ~\infty }\left(1+\sin 4x\right)^{\cot(x)} }\) \(\large\color{#003366}{\displaystyle \ln(y)=\ln\left[\lim_{x \rightarrow ~\infty }\left(1+\sin 4x\right)^{\cot(x)}\right] }\) \(\large\color{#003366}{\displaystyle \ln(y)=\cot(x)\ln\left[\lim_{x \rightarrow ~\infty }\left(1+\sin 4x\right)\right] }\) \(\large\color{#003366}{\displaystyle \ln(y)=\frac{\ln\left[\lim_{x \rightarrow ~\infty }\left(1+\sin 4x\right)\right]}{\tan(x)} }\)
SolomonZelman
  • SolomonZelman
then LHS
zepdrix
  • zepdrix
\[\large\rm \lim_{x\to0^+}\cot x\quad=\lim_{x\to0^+}\frac{\cos x}{\sin x}\]Which is approaching 1/0, yes?
SolomonZelman
  • SolomonZelman
oh 0, I was thinking ∞...
SolomonZelman
  • SolomonZelman
sorry
zepdrix
  • zepdrix
No, infinity is correct :)) The bottom gets really really small, close to 0, so it's blowing up
zepdrix
  • zepdrix
\[\large\rm \lim_{x\to0^+}\frac{\cos x}{\sin x}\quad\approx \frac{1}{(1/99999999)}\quad=99999999\]Take a number really close to 0 for your sin(x), in fraction form it makes a little more sense. 1/99999999 is really close to 0, ya? very tiny decimal value.
SolomonZelman
  • SolomonZelman
\(\large\color{#003366}{ \displaystyle \ln(y)=\displaystyle\lim_{x \rightarrow ~\infty }\frac{\cos x \ln (1+\sin 4x)}{\sin(x)} }\) this is what am arriving at: so it is 0/0
SolomonZelman
  • SolomonZelman
because ln(1)=0
Zenmo
  • Zenmo
Ah, so as the number gets bigger positively, it approaches 0 from the right side. I think I got it now.
Zenmo
  • Zenmo
hence it's positive infinity
SolomonZelman
  • SolomonZelman
wel, the top is also zero, in this case....
SolomonZelman
  • SolomonZelman
not by cos(x)/sin(x), but by: \(\large\color{#003366}{ \displaystyle \ln(y)=\displaystyle\lim_{x \rightarrow ~\infty }\frac{\cos x \ln (1+\sin 4x)}{\sin(x)} }\)
SolomonZelman
  • SolomonZelman
oh sorry
Zenmo
  • Zenmo
Thanks Zepdrix & Soloman :) Fanned you both.
SolomonZelman
  • SolomonZelman
I was thinking 0 now... I keep confusing things
SolomonZelman
  • SolomonZelman
but ∞/∞ ....
SolomonZelman
  • SolomonZelman
(or I would think this is so)
SolomonZelman
  • SolomonZelman
nope it is not ∞, zepdrix is right this imit won't exist.
zepdrix
  • zepdrix
? 0_o I was just referring the first part.. I didn't actually go through the whole thing XD lol
SolomonZelman
  • SolomonZelman
And I forgot the ln just now
SolomonZelman
  • SolomonZelman
with ln it does go to ±∞
SolomonZelman
  • SolomonZelman
\(\large\color{#003366}{ \displaystyle\lim_{x \rightarrow ~\infty }\frac{\cos x \ln (1+\sin 4x)}{\sin(x)} =\frac{-\infty\quad {\rm to}\quad \infty}{0}}\) so far I have worked till this.... ugly enough
SolomonZelman
  • SolomonZelman
http://www.wolframalpha.com/input/?i=lim+x-%3Einfinity+cos%28x%29ln%281%2Bsin%284x%29%29%2Fsin%28x%29 Yes, DNE
zepdrix
  • zepdrix
Solly, you bein silly. I'm pretty sure this one is going to simplify to something like e^4. From here,\[\large\rm \lim_{x\to0^+}\frac{\cos x \ln(1+\sin4x)}{\sin x}\]we want to write it in a clever way,\[\large\rm \lim_{x\to0^+}\frac{\ln(1+\sin4x)}{\frac{\sin x}{\cos x}}\]And now we can see it's approaching 0/0 which is one of our indeterminate forms that allows for the use of L'Hospital's Rule! Yay!
zepdrix
  • zepdrix
Oh you have \(\rm x\to\infty\) in all of your steps... hmm maybe you misread the question :d I'm not sure.
zepdrix
  • zepdrix
You ok with the rest of this one though Zenmo? :) Think you got it?
zepdrix
  • zepdrix
Book prolly details the steps nicely.
SolomonZelman
  • SolomonZelman
Yes, not ∞ my bad, sorry. with ──> it is literally 4
SolomonZelman
  • SolomonZelman
\(\large\color{#003366}{ \displaystyle\lim_{x \rightarrow ~\infty }\frac{\cos x \ln (1+\sin ax)}{\sin(x)} =a}\)
SolomonZelman
  • SolomonZelman
hehe

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