Question

find the limit

Answer 1

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

I got zero using direct substitution but I do not feel that is correct

Answer 2

sorry it is that equation with \[\lim_{x \rightarrow 0}\]

Answer 3

1) Expand the Numerator and Cancel terms:
\[\lim_{x \rightarrow 0} (-\frac{ 1 }{ 2x }+\frac{ 1 }{ x(2+x) })\]
2. After Simplifying:
\[\lim_{x \rightarrow 0} (\frac{ -x }{ 4x+2x^2 })\]
3. Factor out the Constants:
In this case it would be the '-1' which is '-x'
\[- (\lim_{x \rightarrow 0} \frac{ x }{ 4x+2x^2 })\]
4. Factor out the Numerator and the Denominator:
\[- (\lim_{x \rightarrow 0} \frac{ x }{ x(2(2+x)) })\]
5. Now let's assume that x is not equal to 0 and cancel terms:
\[-(\lim_{x \rightarrow 0} \frac{ 1 }{ 2(2+x) })\]
6. Factor out any Constants:
\[-\frac{ 1 }{ 2 }(\lim_{x \rightarrow 0} \frac{ 1 }{ 2+x })\]
7. Simplify:
\[-\frac{ 1 }{ 2(\lim_{x \rightarrow 0} (2+x)}\]
8. Substitute 0 into the equation:
We get 0 from the limit, which is given
\[-(\frac{ 1 }{ 2(2+(0)})\]
9. So your answer will be\[-\frac{ 1 }{ 4 }\]

Answer 4

i dont understand what you did between expanding the numerator and simplifying

Answer 5

do you mean the first two steps?