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Limit with trigonometric and polynomial function.

  1. Oct 23, 2015 #1
    1. The problem statement, all variables and given/known data
    For $$\lim _{ x\rightarrow \infty }{ \frac { { x }^{ 2 }+{ e }^{ -{ x }^{ 2 }\sin ^{ 2 }{ x } } }{ \sqrt { { x }^{ 4 }+1 } } } $$, determine whether it exists. If it does, find its value. if it doesn't, explain.

    2. Relevant equations
    Sand witch theorem and arithmetic rule

    3. The attempt at a solution
    I reached that the limit is 1 using the following:

    $$\lim _{ x\rightarrow \infty }{ \frac { { x }^{ 2 }+{ e }^{ -{ x }^{ 2 }\sin ^{ 2 }{ x } } }{ \sqrt { { x }^{ 4 }+1 } } } =\lim _{ x\rightarrow \infty }{ \frac { { x }^{ 2 } }{ \sqrt { { x }^{ 4 }+1 } } +\frac { 1 }{ { e }^{ { x }^{ 2 }\sin ^{ 2 }{ x } }\sqrt { { x }^{ 4 }+1 } } } \\ \frac { 1 }{ { e }^{ x } } <\frac { 1 }{ { e }^{ { x }^{ 2 } }\sqrt { { x }^{ 4 }+1 } } <\frac { 1 }{ { e }^{ { x }^{ 2 }\sin ^{ 2 }{ x } }\sqrt { { x }^{ 4 }+1 } } <\frac { 1 }{ \sqrt { { x }^{ 4 }+1 } } \\ \\ \lim _{ x\rightarrow \infty }{ \frac { 1 }{ { e }^{ x } } } =\lim _{ x\rightarrow \infty }{ \frac { 1 }{ \sqrt { { x }^{ 4 }+1 } } } =0$$
    Therefore by the sandwich theorem:
    $$\lim _{ x\rightarrow \infty }{ \frac { 1 }{ { e }^{ { x }^{ 2 }\sin ^{ 2 }{ x } }\sqrt { { x }^{ 4 }+1 } } } =0$$

    Also,

    $$\lim _{ x\rightarrow \infty }{ \frac { { x }^{ 2 } }{ \sqrt { { x }^{ 4 }+1 } } } =1$$

    Hence:
    $$\lim _{ x\rightarrow \infty }{ \frac { { x }^{ 2 }+{ e }^{ -{ x }^{ 2 }\sin ^{ 2 }{ x } } }{ \sqrt { { x }^{ 4 }+1 } } } =1+0=1$$

    I tried to put the limit into wolfram but it gave me a time limit exceeded. Is there a reason for this? Does the limit really not exist? And is there anything wrong in my argument?

    Thank you
     
  2. jcsd
  3. Oct 23, 2015 #2

    BvU

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    I think you're doing just fine. which of the two did wolfie suffocate on ?
     
  4. Oct 23, 2015 #3
    Wolfram exceeded time on the original expression without breaking it down. (Although I have no idea why it did)

    And thanks for answering :)
     
  5. Oct 23, 2015 #4

    BvU

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    So you break it down and feed wolfie smaller bites !
     
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