anonymous
  • anonymous
find the limit
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[\frac{ \frac{ 1 }{ 2+x }-\frac{ 1 }{ 2 } }{ x }\] I got zero using direct substitution but I do not feel that is correct
anonymous
  • anonymous
sorry it is that equation with \[\lim_{x \rightarrow 0}\]
anonymous
  • anonymous
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\[-\frac{ 1 }{ 4 }\]

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anonymous
  • anonymous
i dont understand what you did between expanding the numerator and simplifying
anonymous
  • anonymous
do you mean the first two steps?

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