Question

How do you do this limit:
Limit as x approaches 0 of
{1^(1 / sin^2x) + 2^(1 / sin^2 x) + .... + n^(1/ sin^2 x) } ^sin^2 x.
I thought it might be 1 because as x --> 0 then sin^2 x --> 0 but that seems to easy.

Answer 1

It's an interesting question, shouldn't it be infinity?

So you're right on the fact that sin^2 x is 0 when x=0. So simplifying that it would give

1^0 + 2^0 + 3^0 ....

So that's all 1, so you're having an arithmetic series of 1 adding up, and it supposedly does not converge. So it's just infinity, but I'm not sure if I'm wrong.

Answer 2

Oh wait, I did not see the whole thing raised to another sin^2 x.

Then I guess you are right because it's all raised to sin^2 x which when x approaches 0, it's 1.

Sometimes they give questions that are too long but require 1 step. I've had those times too, just staring at the question for minutes.

Answer 3

What's the answer? Is it different?

Answer 4

I don't know the answer. I played around with some numbers though.
If we make x small say 0.1 radians. then 2^ (1 / sin^2x) = about 1.598 * 10^30 . A huge number ! Now if you take that to the power sin^0.1 it works out to 2!. Then I tried the same calculation with 2^(1/ sin^20.1) + 3^(1 / sin^2 0.1) and it work out to 3. So maybe the anser is n???

Answer 5

The first term is 1 of course but the higher terms work out as enormous numbers I tried working out the 5th term but the calculator couldn't handle the huge number. - and 0.1 radians is not a very small value.

Answer 6

Limits can be tricky. I expect we could use a substitution here but can't think of an appropriate one.

Answer 7

@Hero can you help with this?

Answer 8

or @Vocaloid?

Answer 9

maybe the Question is designed to trick u...

Answer 10

btw \[\large \rm 1^\frac{ 1 }{ \sin^2(x) } \cancel{=} 1^0\]

Answer 11

\[\large \rm 1^\frac{ 1 }{ \sin^2(x) } = 1^\frac{1}{0}\]

Answer 12

\[\Large \rm (1^\frac{ 1 }{ \sin^2(x) }+2^\frac{ 1 }{ \sin^2 (x)}+....n^\frac{ 1 }{ \sin^2(x)})^\color{Red}{{\sin^2(x)}}\] is this correct??

Answer 13

\[\rm (anyting)^\color{Red}{\sin^2(0)}--> (anything)^\color{Red}{0}=1\]

Answer 14

Yes that was my initial thought on the question.

Answer 15

Yes i was wrong about the first term being = 1.

Answer 16

According to someone on Brainly the answer is n. I've lost the post to give the details. I'll look again.

Answer 17

ok take a screenshot and post it here will see.

Answer 18

Don't know why it is so small .

Answer 19

You can just click on it to get a larger view. I'm afraid some of that math is beyond me.

Answer 20

well for your question x approaches 0 not infinity.

Answer 21

I'll check that out.

Answer 22

Yes . The original question wanted the limit to 0, so the answerer misread the question.

Answer 23

yeah. so when x approaches infinity the exponents sin2(x) will not be 0 anymore so the answer will be diff.

Answer 24

OK Thanks.

Answer 25

np :=))

Answer 26

I wonder if there is another way to solve this where we don't get 1/0 exponent or maybe we just have to ingore everything since it rasied to the power of sin^2(x) . also it really depends on the level of math u r taking.

Answer 27

I guess this one is degree level . I'm educated to advanced level (UK). I just came across this one in Brainly and found it interesting. I wonder if Wolframalpha could do this one????

Answer 28

https://www.wolframalpha.com/input/?i=limit+as+x+approaches+0+of+%281%5E%281%2Fsin%5E2%28x%29%29%2B2%5E%281%2Fsin%5E2%28x%29%29%2B3%5E%281%2Fsin%5E2%28x%29%29%2B....%2Bn%5E%281%2Fsin%5E2%28x%29%29%29%5E%28sin%5E2%28x%29%29
I hope i typed it correclty

Answer 29

Yes it is correct . Answer 1!

Answer 30

https://www.wolframalpha.com/input/?i=limit+as+x+approaches+0+of+%281%5E%281%2Fsin%5E2%28x%29%29%2B2%5E%281%2Fsin%5E2%28x%29%29%2B3%5E%281%2Fsin%5E2%28x%29%29%29%5E%28sin%5E2%28x%29%29