Limit with trigonometric and polynomial function.

[FaroukYasser]

Homework Statement

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.

Homework Equations

Sand witch theorem and arithmetic rule

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

 

[BvU]

I think you're doing just fine. which of the two did wolfie suffocate on ?

 

[FaroukYasser]

I think you're doing just fine. which of the two did wolfie suffocate on ?

Wolfram exceeded time on the original expression without breaking it down. (Although I have no idea why it did)

And thanks for answering :)

 

[BvU]

So you break it down and feed wolfie smaller bites !