Question

Calculate limits (Question on indeterminate forms).

Answer 1

\[\lim_{x \rightarrow 0^+}(1+\sin4x)^{cotx}\] How is 1+sin4x --> 1 and cotx --> infinity? In order to know if the given limit is indeterminate.

Answer 2

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.

Answer 3

Yea, I know how to solve it. But, I have to check if its indeterminate or not. Thanks for trying though. :)

Answer 4

that indeterminate form doesn't seem to do much to me the way the limit is now...

Answer 5

I'm reading a book, trying to make a sense out of it.

Answer 6

But 1+sin(4x) is not 1

Answer 7

That is 0 to 2, I checked via wolfr

Answer 8

so 2^∞=∞, 0^∞=0

Answer 9

? 0_o

At x=0: 1+sin(4x) = 1, ya? :o

Answer 10

\(\large\color{#003366}{\displaystyle y=\lim_{x \rightarrow ~\infty }\left(1+\sin 4x\right)^{\cot(x)} }\)

ln both sides, tnx for the link :)

Answer 11

ok so x=0: 1+sin(4x)=1. I got that, but \[\cot(x)=\]?

Answer 12

cot(x)=positive infinity*

Answer 13

\(\large\color{#003366}{\displaystyle y=\lim_{x \rightarrow ~\infty }\left(1+\sin 4x\right)^{\cot(x)} }\)

\(\large\color{#003366}{\displaystyle \ln(y)=\ln\left[\lim_{x \rightarrow ~\infty }\left(1+\sin 4x\right)^{\cot(x)}\right] }\)

\(\large\color{#003366}{\displaystyle \ln(y)=\cot(x)\ln\left[\lim_{x \rightarrow ~\infty }\left(1+\sin 4x\right)\right] }\)

\(\large\color{#003366}{\displaystyle \ln(y)=\frac{\ln\left[\lim_{x \rightarrow ~\infty }\left(1+\sin 4x\right)\right]}{\tan(x)} }\)

Answer 14

then LHS

Answer 15

\[\large\rm \lim_{x\to0^+}\cot x\quad=\lim_{x\to0^+}\frac{\cos x}{\sin x}\]Which is approaching 1/0, yes?

Answer 16

oh 0, I was thinking ∞...

Answer 17

sorry

Answer 18

No, infinity is correct :))
The bottom gets really really small, close to 0, so it's blowing up

Answer 19

\[\large\rm \lim_{x\to0^+}\frac{\cos x}{\sin x}\quad\approx \frac{1}{(1/99999999)}\quad=99999999\]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.

Answer 20

\(\large\color{#003366}{ \displaystyle \ln(y)=\displaystyle\lim_{x \rightarrow ~\infty }\frac{\cos x \ln (1+\sin 4x)}{\sin(x)} }\)

this is what am arriving at:

so it is 0/0

Answer 21

because ln(1)=0

Answer 22

Ah, so as the number gets bigger positively, it approaches 0 from the right side. I think I got it now.

Answer 23

hence it's positive infinity

Answer 24

wel, the top is also zero, in this case....

Answer 25

not by cos(x)/sin(x),

but by:

\(\large\color{#003366}{ \displaystyle \ln(y)=\displaystyle\lim_{x \rightarrow ~\infty }\frac{\cos x \ln (1+\sin 4x)}{\sin(x)} }\)

Answer 26

oh sorry

Answer 27

Thanks Zepdrix & Soloman :) Fanned you both.

Answer 28

I was thinking 0 now... I keep confusing things

Answer 29

but ∞/∞ ....

Answer 30

(or I would think this is so)

Answer 31

nope it is not ∞, zepdrix is right this imit won't exist.

Answer 32

? 0_o
I was just referring the first part.. I didn't actually go through the whole thing XD lol

Answer 33

And I forgot the ln just now

Answer 34

with ln it does go to ±∞

Answer 35

\(\large\color{#003366}{ \displaystyle\lim_{x \rightarrow ~\infty }\frac{\cos x \ln (1+\sin 4x)}{\sin(x)} =\frac{-\infty\quad {\rm to}\quad \infty}{0}}\)

so far I have worked till this....
ugly enough

Answer 36

http://www.wolframalpha.com/input/?i=lim+x-%3Einfinity+cos%28x%29ln%281%2Bsin%284x%29%29%2Fsin%28x%29

Yes, DNE

Answer 37

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!

Answer 38

Oh you have \(\rm x\to\infty\) in all of your steps...
hmm maybe you misread the question :d I'm not sure.

Answer 39

You ok with the rest of this one though Zenmo? :)
Think you got it?

Answer 40

Book prolly details the steps nicely.

Answer 41

Yes, not ∞ my bad, sorry.
with ──> it is literally 4

Answer 42

\(\large\color{#003366}{ \displaystyle\lim_{x \rightarrow ~\infty }\frac{\cos x \ln (1+\sin ax)}{\sin(x)} =a}\)

Answer 43

hehe