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### Using L'Hopital's Rule

```
Date: 01/22/99 at 05:17:12
From: Mirza
Subject: What is the limit of ....

Could you give me the proof that e^(-1/3) is the limit for

y = (arctan(x)/x)^(1/(x^2))

when x -> 0?
```

```
Date: 01/22/99 at 08:48:31
From: Doctor Anthony
Subject: Re: What is the limit of ....

Take logs:

ln(y) = (1/x^2)ln[tan^(-1)(x)/x]

ln[tan^(-1)(x)/x]
ln(y) = -----------------
x^2

Note that tan^(-1)(x)/x  -> 1 as x -> 0 and ln(1) = 0. So the righthand
side tends to the value 0/0 as x -> 0, and we can apply l'Hopital's
rule. Differentiating the top and bottom lines separately we have:

[x/tan^(-1)(x)][x/(1+x^2)-tan^(-1)(x)]/x^2
---------------------------------------------
2x

x - tan^(-1)(x) (1+x^2)
= ------------------------
2x^2 tan^(-1)(x) (1+x^2)

As x -> 0, this is still 0/0, so we repeat the process. Differentiating
the top and bottom separately

1 - [(1+x^2)/(1+x^2) + tan^(-1)(x) 2x]
--------------------------------------------
2(x^2+x^4)/(1+x^2) + 2 tan^(-1)(x) (2x+4x^3)

which simplifies to

-tan^(-1)(x)
-------------------------
x + 2(1+2x^2)tan^(-1)(x)

This is still 0/0 so repeat the process. Differentiate:

-1/(1+x^2)
--------------------------------------
1 + 2(1+2x^2)/(1+x^2) + 8x tan^(-1)(x)

Then multiply top and bottom by (1+x^2):

-1
------------------------------------------
1 + x^2 + 2(1+2x^2) + 8x(1+x^2)tan^(-1)(x)

Putting x = 0, this reduces to:

-1      -1
----- =  ---
1 + 2     3

So the limit of ln(y) = -1/3. Thus the limit of y = e^(-1/3).

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Analysis
High School Calculus

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