Question

could you guys help me with this limit please only using algebra techniques.

Answer 1

\[\lim_{x \rightarrow 0^+}\frac{\sin^{-1}( \sqrt{x})}{1-\cos(4 \sqrt{x})}\]

Answer 2

yeah

Answer 3

so pluggin in 0 won't work because you would get 0/(1-1)=0/0
so this means there is more work to do

Answer 4

Ok I'm not totally sure this will work but we can try multiplying the conjugate of the bot on top and bot

Answer 5

by 1 + cos 4 sqrt (x)

Answer 6

so when i multiply 1 - cos 4 sqrt(x) * by 1 + cos 4 sqrt (x) = 1 - cos^2 16(x) is that right ?

Answer 7

so that really is arcsin on top right?

Answer 8

yeah arcsin

Answer 9

any special limits you have learned that involve arcsin

Answer 10

not really

Answer 11

i was just curious about that limit saw on sao tan calculus book

Answer 12

it*

Answer 13

yeah just curious to solve it

Answer 14

and it said just to use algebraic techniques?

Answer 15

well in my exams im only allowed to use algebra techniques

Answer 16

I can use the squeeze theroem

Answer 17

in my limit exam*

Answer 18

Is squeeze theorem allowed?

Answer 19

yeah

Answer 20

also called sandwich theorem

Answer 21

Great!

Answer 22

yeah absolutely

Answer 23

arcsin is between -pi/2 and pi/2

Answer 24

like you will only ever get an output between -pi/2 and pi/2 (inclusive)

Answer 25

for arcsin

Answer 26

yeah right

Answer 27

so take that interval as reference ?

Answer 28

\[\frac{-\frac{\pi}{2}}{1-\cos(4 \sqrt{x})} \le \frac{\arcsin(x)}{1-\cos(4 \sqrt{x})} \le \frac{\frac{\pi}{2}}{1-\cos(4 \sqrt{x})}\]

Answer 29

whoa

Answer 30

but limit in den will still be 0

Answer 31

\[\lim_{x \rightarrow 0^+}\frac{\frac{\pi}{2}}{1-\cos(4 \sqrt{x})}\]
we know this will be infinity/-infinity/does not exist

Answer 32

one of those 3

Answer 33

so if equation on the sides tend to infiny the limit of the limit is infinity right?

Answer 34

the middle limit will be infinity as well

Answer 35

the only thing is i don't think this will work
since the limit on the right goes to infinity
and the limit on the left goes to negative infinity

Answer 36

:(

Answer 37

oh i see

Answer 38

heheh well kind of diffucult limit isn't

Answer 39

i save it for later

Answer 40

@amistre64 @zepdrix

Answer 41

I'm kinda stumped right now

Answer 42

@thomaster @iambatman any ideas?

Answer 43

\[\lim_{x \rightarrow 0}\frac{\cos(x)-1}{x^2} \cdot \frac{\cos(x)+1}{\cos(x)+1}\]

Answer 44

on top what do you get ?

Answer 45

hmm sin x ?

Answer 46

sin^2 x

Answer 47

and sin(x)/x goes to what as x goes to 0?

Answer 48

@amistre64 I wanted you to look at this:
\[\lim_{x \rightarrow 0^+}\frac{\sin^{-1}( \sqrt{x})}{1-\cos(4 \sqrt{x})} \]

Answer 49

oh yeah

Answer 50

\[\lim_{x \rightarrow 0}\frac{\sin(x)}{x}=1 \\ \lim_{x \rightarrow 0}\frac{\sin(x)}{x} \frac{\sin(x)}{x}=1 \cdot 1 =1\]

Answer 51

\[\lim_{x \rightarrow 0}\frac{\cos(x)-1}{x^2} \cdot \frac{\cos(x)+1}{\cos(x)+1} =\lim_{x \rightarrow 0}\frac{-\sin^2(x)}{x} \frac{1}{\cos(x)+1}\]
Do you think you can finish this one?

Answer 52

im on it

Answer 53

y = sin^-1(sqrt(x))

sin(y) = sqrt(x) wonder if this is useful

Answer 54

and i left the square off the in the denominator by the way @Bryan11

Answer 55

\[\lim_{x \rightarrow 0}\frac{\cos(x)-1}{x^2} \cdot \frac{\cos(x)+1}{\cos(x)+1} =\lim_{x \rightarrow 0}\frac{-\sin^2(x)}{x^2} \frac{1}{\cos(x)+1}\]

Answer 56

hey i got 1/2 is that right?

Answer 57

there should be a negative in front of it

Answer 58

ohh right yeah my mistake

Answer 59

And on that previous one @amistre64 wants to try to make a sub

Answer 60

so in this expression sin^2x / x^2 i know it looks like sinx / x but how do i explain it step by step

Answer 61

sin^2(x)/x^2=sin(x)/x*sin(x)/x

Answer 62

is trig considered algebra?

Answer 63

good question

Answer 64

yeah trig can be used i think

Answer 65

identities and all that stuff

Answer 66

i think i see some squeeze thrms being tried in this

Answer 67

I tried squeeze theorem but I think I selected the wrong functions

Answer 68

we know the limit tho dont we .... wolf says it +inf i think

Answer 69

yah

Answer 70

oh the limit is really infinity

Answer 71

so you were right

Answer 72

since goes to 0 from the right maybe we could have chosen
so arcsin will be between 0 and 1
\[\frac{?}{1-\cos(4 \sqrt{x})} \le \frac{\sin^{-1}(\sqrt{x})}{1-\cos(4 \sqrt{x})} \le \frac{1}{1-\cos(4 \sqrt{x})}\]

Answer 73

but I don't want the ? to mark to be 0

Answer 74

maybe the ? can be sqrt(x) or x
since as this inequality would be true for values approaching 0 from the right

Answer 75

why is not -1

Answer 76

\[\frac{ \sqrt{x}}{1-\cos(4 \sqrt{x})} \le \frac{\sin^{-1}(\sqrt{x})}{1-\cos(4 \sqrt{x})} \le \frac{1}{1-\cos(4 \sqrt{x})}\]

Answer 77

errr... one sec I still messed this inequality up

Answer 78

\[\frac{ \sqrt{x}}{1-\cos(4 \sqrt{x})} \le \frac{\sin^{-1}(\sqrt{x})}{1-\cos(4 \sqrt{x})} \le \frac{\frac{\pi}{2}}{1-\cos(4 \sqrt{x})}\]

Answer 79

ok though that one still goes to infinity (they actually all go to infinity)

Answer 80

So we are looking arcsin(sqrt(x)) between 0 and pi/2
because x is suppose to be to the right of 0

Answer 81

sqrt(x) goes to 0 so instead of putting 0 on top of that fraction I put sqrt(x) on top of that one fraction

Answer 82

oh i see

Answer 83

yeah i think that's it

Answer 84

You re really good tank so much

Answer 85

hey may i ask you something, can you recommend me some really calculus book to get exercises on limits

Answer 86

already did steward's and thomas calculus

Answer 87

I have a steward calculus 6th edition

Answer 88

and that is the only cal book i have

Answer 89

but it always happend to me that the teacher pulls out some really hard exercises on the test

Answer 90

But I could try to find you a good website

Answer 91

that was a difficult question

Answer 92

@amistre64 are you fine with my inequality?

Answer 93

hehee i mean the teacher always puts difficult exercises on the test and they dont even appear on any text book

Answer 94

I think I'm happy with my inequality
You just have to show that left one also goes to infinity which it does