how to find lim x → 0 sin 2 ⁡ ( 1 x ) sin 2 ⁡ x

Question

how to find       lim          x      →      0            sin    2    ⁡  (      1    x    )      sin    2    ⁡  x

Anika Stevenson · Accepted Answer

Just notice that        −  1  ≤  sin  ⁡  t  ≤  1      ⟹      0  ≤      sin    2    ⁡      (          1      x        )        sin    2    ⁡  x  ≤      sin    2    ⁡  xAnd then use the squeeze theorem.

Quintin Stafford · Accepted Answer

its hint for first answerSqueeze theorem:  0  ≤      sin    2    ⁡  (      1    x    )  ≤  1  ⇒  0  ≤      sin    2    ⁡  (      1    x    )      sin    2    ⁡  (  x  )  ≤      sin    2    ⁡  (  x  )And      lim          x      →      0            sin    2    ⁡  (  x  )  =  0  =      lim          x      →      0        0So      lim          x      →      0            sin    2    ⁡  (      1    x    )      sin    2    ⁡  (  x  )  =  0

how to find <munder> <mo movablelimits="true" form="prefix">lim <mrow class="MJX-TeXAt

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