Say I have to calculate the limit <munder> <mo movablelimits="true" form="prefix">lim <m

Simone Werner

Simone Werner

Answered question

2022-05-24

Say I have to calculate the limit
lim x 0 1 cos ( x ) 3 sin 2 ( x )

Answer & Explanation

pralkammj

pralkammj

Beginner2022-05-25Added 7 answers

We have to show that
ϵ > 0 : δ ( 0 , π / 2 ] : x R { 0 } : ( | x | < δ | f ( x ) | < ϵ )
where
f ( x ) = ( 1 cos x ) 2 6 sin 2 x = ( 1 cos x ) 2 6 ( 1 cos 2 x )
Let t = cos x. Then, because t 1 whenever x 0 and | x | < π / 2
f ( x ) = ( 1 t ) 2 6 ( 1 t 2 ) = ( 1 t ) 6 ( 1 + t ) = ( 2 1 t ) 6 ( 1 + t ) = 1 3 ( 1 + t ) 1 6
f(x) strictly decreases w.r.t. t
Find t that makes f ( x ) = ϵ. After all, t = 2 / ( 6 ϵ + 1 ) 1. Since t = cos x, choosing
δ = arccos ( 2 6 min { ϵ , 1 / 6 } + 1 1 )
proves the claim. I think you can show that δ is an increasing function of ϵ
Brennen Fisher

Brennen Fisher

Beginner2022-05-26Added 3 answers

You can also multiply and divide by 1 + cos x
lim x 0 1 cos x 3 sin 2 x = lim x 0 1 cos 2 x 3 sin 2 x ( 1 + cos x ) = lim x 0 sin 2 x 3 sin 2 x ( 1 + cos x ) = 1 3 lim x 0 1 1 + cos x = 1 6
If you need to use the definition, you can easily see that
| 1 cos x 3 sin 2 x 1 6 | = 1 3 | 1 1 + cos x 1 2 | 0 ( x 0 )
Since there is no indetermination in this limit, you can use Heine's definition in a very straightforward way (instead of Cauchy's). Or, you can go on to obtain
| 1 1 + cos x 1 2 | = 1 2 | 1 cos x 1 + cos x | 1 2 | 1 cos x | 1 2 | x |

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