Question

help solving a limit where x approaches 0
Need help solving algebraically

Answer 1

The equation is \[\lim_{x \rightarrow 0} \frac{ \frac{ 1 }{ 2+x }-\frac{ 1 }{ 2 } }{ x }\]

Answer 2

how would i solve this algebraically?

Answer 3

Well focus on the numerator first

\[\large \frac{1}{2 + x} - \frac{1}{2}\]

What is the common denominator here?

Answer 4

2

Answer 5

Well actually is would be 2(x + 2)

If we multiply the first fraction by 2...and the second fraction by (2 + x) we will have

\[\large \frac{2}{2(2 + x)} - \frac{2 + x}{2(2 + x)}\]

Now we can put them over the common denominator

\[\large \frac{2 - (2 + x)}{2(2 + x)}\]

simplify that a bit more...

Answer 6

2-1 over 2? so 1/2?

Answer 7

\[\large \frac{2 - (2 + x)}{2(x + 2)}\]
distribute out the '-' sign
\[\large \frac{2 - 2 - x}{2(x + 2)}\]

so

\[\large \frac{-x}{2(x + 2)}\]

make sense there?

well that was just the numerator of the original question

So lets add in the 'x' we ignored before

\[\Large \frac{\frac{-x}{2(x + 2)}}{x}\]

what do we do next?

Answer 8

so u multiply by 1/x to the whole thing making it \[\frac{ -x }{ 2x(2+x) }\]

Answer 9

then the x cancels out, making it \[\frac{ -1 }{ 2(2+x) }\]

Answer 10

and by substituting, u get -1/4 correct?

Answer 11

@johnweldon1993 is that correct?

Answer 12

Correct indeed