Lim x -> 0 of (1/(2+x) - 1/2)/x please help!

  • #1
217
0
lim x tends to 0 of:

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

I simplified the fraction and got:

[x / 2(2+x)] / x

and with further simplification:

x / x(4+2x)

How do I solve this without using L'Hopital's rule?
 

Answers and Replies

  • #2
217
0
The answer is 1, by the way.
 
  • #3
You can use "partial fractions" to simplify.
[1/(2+x) - 1/2] / x <=> 1/x(2+x) - 1/2x <=> 1/2x-1/(2x+4)-1/2x <=> -1/(2x+4)

When x goes to 0, the solution is goes towards -1/4 = -1/(2*0+4)

The partial fraction of found by solving
1/x(2+x) = a/x + b/(x+2)
 
  • #4
You can use "partial fractions" to simplify.
[1/(2+x) - 1/2] / x <=> 1/x(2+x) - 1/2x <=> 1/2x-1/(2x+4)-1/2x <=> -1/(2x+4)

When x goes to 0, the solution is goes towards -1/4 = -1/(2*0+4)

The partial fraction of found by solving
1/x(2+x) = a/x + b/(x+2)
 

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