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$${\frac{\left({\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{x}}}}\right)}{\mathtt{\,-\,}}{\mathtt{e}}\right)}{{\mathtt{x}}}}$$Programmer

difficulty advanced
Guest Jun 22, 2015

Best Answer 

 #1
avatar+26322 
+10

Limit

.

Alan  Jun 22, 2015
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8+0 Answers

 #1
avatar+26322 
+10
Best Answer

Limit

.

Alan  Jun 22, 2015
 #2
avatar+90988 
+5

Thank you Alan,

How did you do the initial expansion?  What kind of an expansion is that?

Melody  Jun 22, 2015
 #3
avatar+26322 
+5

I just used Mathcad to tell me Melody!  I'll see if I can derive it.

.

Alan  Jun 22, 2015
 #4
avatar+90988 
0

Thanks Alan but worry too much, I don't understand expansions much.   :(

I only know the binomial expansion.

I just wondered :/

Melody  Jun 22, 2015
 #5
avatar+26322 
+5

Here's a derivation explicitly using the binomial expansion:

 

 limit:

.

Alan  Jun 22, 2015
 #6
avatar+90988 
0

Thanks Alan,

I can follow most of that but you lost me at     $$O(x^2)$$     and what happened to all the other powers of x?

 

Only answer is you feel enthused because this stuff is really over my head.  I couldn't reproduce it.  :(

Melody  Jun 22, 2015
 #7
avatar+26322 
+5

Rather than writing all the other terms involving x2x3 etc. I've written O(x2), meaning terms of order x2 and smaller ("smaller" because we are going to make x go to 0, so x4, x3 etc. are going to be smaller than x2). The important thing is that all these other terms will be multiples of xn, where n>1, so that when they are divided by x they are still mutiples of powers of x and hence vanish in the limit as x tends to 0.

.

Alan  Jun 22, 2015
 #8
avatar+90988 
0

Thank you Alan,

I shall think on this :)

Melody  Jun 22, 2015

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