Using the Squeeze Theorem to evaluate lim x → 0 x 3 cos ⁡ ( ln ⁡ ( x 4 ) )Would this just be a simple application of the Squeeze Theorem? (I can't use L'hopital or Taylor polynomials.) − x 3 ≤ x 3 cos ⁡ ( ln ⁡ ( x 4 ) ) ≤ x 3 The limit from both sides is 0, so the middle limit must also be 0. Is that right?

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Using the Squeeze Theorem to evaluate       lim          x      →      0            x    3    cos  ⁡  (  ln  ⁡  (      x    4    )  )Would this just be a simple application of the Squeeze Theorem? (I can&#039;t use L&#039;hopital or Taylor polynomials.)  −      x    3    ≤      x    3    cos  ⁡  (  ln  ⁡  (      x    4    )  )  ≤      x    3  The limit from both sides is 0, so the middle limit must also be 0. Is that right?

Jake Mcpherson · Accepted Answer

Yes, you are right. Since  −  1  ≤  cos  ⁡  (  ln  ⁡  (      x    4    )  )  ≤  1for all x (  ≠  0) you indeed have  −      x    3    ≤      x    3    cos  ⁡  (  ln  ⁡  (      x    4    )  )  ≤      x    3  and so      lim          x      →      0        −      x    3    =      lim          x      →      0            x    3    =  0by the Squeeze Theorem the function in the middle will also have limit 0:      lim          x      →      0            x    3    cos  ⁡  (  ln  ⁡  (      x    4    )  )  =  0.

Using the Squeeze Theorem to evaluate <munder> <mo movablelimits="true" form="prefix">lim

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