Limiting x->∞: (e^(x^-2)-1)/(π-2arctan(x^2))

  • Thread starter Tanishq Nandan
  • Start date
In summary: Yep,-1/2 (I used another inverse property in b/w though,I had gone wrong in writing the last step,that way I didn't need l'hospital)
  • #1
Tanishq Nandan
122
5

Homework Statement



Find limit for x tending to infinity of: (e^(x^-2)-1)/(2arctan(x^2) - pi)

Homework Equations


arctanx +arccotx=pi/2
arctanx=arccot(1/x)

The Attempt at a Solution


20170618_003226-1.jpg

P.S I haven't written limit before each step, and hope you can get how I arrived at the first step.I just took 2 common from the denominator and used simple inverse trig.

However,the answer is -1/2
Where am I going wrong?
 
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  • #2
##\frac{1}{\cot^{-1}(x^2)} = \tan^{-1}(x^2)## is not true.

I didn't check whether you made other mistakes though.
 
  • #3
Tanishq Nandan said:

Homework Statement



Find limit for x tending to infinity of: (e^(x^-2)-1)/(2arctan(x^2) - pi)

Homework Equations


arctanx +arccotx=pi/2
arctanx=arccot(1/x)

The Attempt at a Solution


View attachment 205654
P.S I haven't written limit before each step, and hope you can get how I arrived at the first step.I just took 2 common from the denominator and used simple inverse trig.

However,the answer is -1/2
Where am I going wrong?
Do you know L'Hopital's Rule?
 
  • #4
Math_QED said:
##\frac{1}{\cot^{-1}(x^2)} = \tan^{-1}(x^2)## is not true.

I didn't check whether you made other mistakes though.
Ok
 
  • #5
Yes,I do
Mark44 said:
Do you know L'Hopital's Rule?
 
  • #6
Ok,I got where I am going wrong,but how to go right?
It looks like we won't be able to use inverse trig
 
  • #7
Tanishq Nandan said:
Ok,I got where I am going wrong,but how to go right?
It looks like we won't be able to use inverse trig

Use l'Hopital's rule on the first step you wrote down and you should be fine.
 
  • #8
Tanishq Nandan said:
Ok,I got where I am going wrong,but how to go right?
It looks like we won't be able to use inverse trig
Can you find the derivative of ##\ \tan^{-1}(x^2) - \frac \pi 2 \,,\ ## and derivative of ##\ e^{1/x^2}-1\,?##
 
  • #9
Yeah,I can..and I did,thanks
 
  • #10
Tanishq Nandan said:
Yeah,I can..and I did,thanks

Did you get the right answer?
 
  • #11
Yep,-1/2 (I used another inverse property in b/w though,I had gone wrong in writing the last step,that way I didn't need l'hospital)
 

1. What does the expression (e^(x^-2)-1)/(π-2arctan(x^2)) approach as x approaches infinity?

The expression approaches 0 as x approaches infinity.

2. Is this expression defined for all real values of x?

No, this expression is undefined for x = 0.

3. How does the graph of (e^(x^-2)-1)/(π-2arctan(x^2)) behave as x approaches infinity?

The graph approaches the horizontal line y=0 as x approaches infinity.

4. Can this expression be simplified?

Yes, it can be simplified to e^(-2x)/π as x approaches infinity.

5. Are there any other values that the expression approaches besides 0?

No, as x approaches infinity, the only value that the expression approaches is 0.

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