I'm lost on the following problem. I don't know which method to use. <munder> <mo movablelim

Jaeden Weaver

Jaeden Weaver

Answered question

2022-05-11

I'm lost on the following problem. I don't know which method to use.
lim x sin ( 2 x ) + 2 x sin ( 1 x )

Answer & Explanation

erlanpb8cj

erlanpb8cj

Beginner2022-05-12Added 14 answers

You can solve the limit without using a double-angle identity:
lim x sin ( 2 / x ) + 2 / x sin ( 1 / x ) = lim t 0 sin ( 2 t ) + 2 t sin ( t ) = lim t 0 [ sin ( 2 t ) sin ( t ) + 2 t sin ( t ) ] = lim t 0 [ sin ( 2 t ) 1 1 sin ( t ) + 2 t sin ( t ) ] = lim t 0 [ sin ( 2 t ) 2 t 2 t sin ( t ) + 2 t sin ( t ) ] = lim t 0 sin ( 2 t ) 2 t 2 lim t 0 t sin ( t ) + 2 lim t 0 t sin ( t ) = 1 2 + 2 = 4

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