anonymous
  • anonymous
Help finding the limit of a function.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Alienium
  • Alienium
ask...
anonymous
  • anonymous
\[\lim_{x \rightarrow 0}\frac{ \frac{ 1 }{ 2+x }-\frac{ 1 }{ 2 } }{ x }\]
Luigi0210
  • Luigi0210
Well..

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anonymous
  • anonymous
I can't just substitute in zero.
Alienium
  • Alienium
solve lim x -> 0 of 1/2+z " " " " " " " - 1/2 take x -> 0 get the trend of values
anonymous
  • anonymous
Wouldn't the limit as x->0 of 1/2+x just be 1/2?
Alienium
  • Alienium
oh... a nice trick... the lim of a simple equation is near the value of the equation when the x = 0 so substitute x = 0 and the value is the limit
anonymous
  • anonymous
I would get 0/0= 0 My book has something else.
Alienium
  • Alienium
oooh so it's a special case 0/0 those have some special solutions... i don't recall right now
anonymous
  • anonymous
That's alright. @dan815
anonymous
  • anonymous
\[\begin{align*}\lim_{x\to0}\frac{\dfrac{1}{2+x}-\dfrac{1}{2}}{x}&=\lim_{x\to0}\frac{\dfrac{2}{2(2+x)}-\dfrac{2+x}{2(2+x)}}{x}\\ &=\lim_{x\to0}\frac{\dfrac{x}{2(2+x)}}{x}\\\\\\ &=\lim_{x\to0}\dfrac{1}{2(2+x)}\\ &=\cdots \end{align*}\]
anonymous
  • anonymous
Thank you so much. I was thinking about doing that, but when I was doing it mentally, I still got zero for the numerator. I should have actually worked it out... Thanks again!

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