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## #1 2017-08-02 17:41:25

NakulG
Member
Registered: 2014-09-02
Posts: 186

### Limits Difficult

Can anyone help solve this limit.

Thanks

Lim x-->0,  [1 -  (x^2)/2) - Cos ( x / (1 - x^2 ) ) ] / x^4

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## #2 2017-08-02 20:18:53

zetafunc
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Registered: 2014-05-21
Posts: 2,210
Website

### Re: Limits Difficult

Which of these is your limit?

The first limit gives
, so I suspect you mean the lower one? If so, it looks like four applications of L'Hopital would do it (Mathematica gives 23/24 as the solution, which I assume comes from the repeated differentiation) but this seems far too tedious. I will see if there is a simpler way.

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## #3 2017-08-02 20:28:49

NakulG
Member
Registered: 2014-09-02
Posts: 186

### Re: Limits Difficult

yes the lower one

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## #4 2017-08-04 01:23:33

zetafunc
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Registered: 2014-05-21
Posts: 2,210
Website

### Re: Limits Difficult

Okay, here's a quicker way to do it, without using L'Hopital's rule four times -- we'll use the Taylor series for
, i.e. we use

Just replace the cosine term with that -- you can disregard the higher order term, because that'll just tend to zero in the limit since the denominator is
. The
s cancel, so you can cancel out an
term from the numerator and denominator, and you should be able to do the rest from there. Let me know if you have any more problems.

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## #5 2017-08-07 18:09:41

NakulG
Member
Registered: 2014-09-02
Posts: 186

### Re: Limits Difficult

Got it thanks
Taylor Series --> Maclaurin Expansion

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## #6 2017-08-08 04:39:19

zetafunc
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Registered: 2014-05-21
Posts: 2,210
Website

### Re: Limits Difficult

You're welcome! Yes, that is the Maclaurin series for cosine. Math Is Fun has its own page on Taylor series here.

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