Help finding the limit of a function.

Question

alienium · Answer

ask...

anonymous · Answer

$$\lim_{x \rightarrow 0}\frac{ \frac{ 1 }{ 2+x }-\frac{ 1 }{ 2 } }{ x }$$

luigi0210 · Answer

Well..

anonymous · Answer

I can't just substitute in zero.

alienium · Answer

solve lim x -> 0 of 1/2+z
"          "   "  "    "  "      "       -  1/2
take x -> 0 get the trend of values

anonymous · Answer

Wouldn't the limit as x->0 of 1/2+x just be 1/2?

alienium · Answer

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 · Answer

I would get 0/0= 0 My book has something else.

alienium · Answer

oooh so it's a special case 0/0
those have some special solutions... i don't recall right now

anonymous · Answer

That's alright. @dan815

anonymous · Answer

$$\begin{align*}\lim_{x\to0}\frac{\dfrac{1}{2+x}-\dfrac{1}{2}}{x}&=\lim_{x\to0}\frac{\dfrac{2}{2(2+x)}-\dfrac{2+x}{2(2+x)}}{x}\
&=\lim_{x\to0}\frac{\dfrac{x}{2(2+x)}}{x}\\\
&=\lim_{x\to0}\dfrac{1}{2(2+x)}\
&=\cdots 
\end{align*}$$

anonymous · Answer

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!