# How do you evaluate \lim _ { x \rightarrow 2} \frac { \frac { 1} { x } - \frac { 1} { 2} } { x - 2}?

##### 2 Answers
Oct 22, 2017

$- \setminus \frac{1}{4}$

#### Explanation:

Using L'Hôpital's rule, we can differentiate both numerator and denominator, and the limit in question is equal to the new limit, that is:

$\setminus {\lim}_{x \setminus \rightarrow 2} \setminus \frac{\setminus \frac{1}{x} - \setminus \frac{1}{2}}{x - 2} = \setminus {\lim}_{x \setminus \rightarrow 2} \setminus \frac{- \setminus \frac{1}{{x}^{2}} - 0}{1 - 0} = \setminus {\lim}_{x \setminus \rightarrow 2} \left(- \setminus \frac{1}{{x}^{2}}\right) = - \setminus \frac{1}{{2}^{2}} = - \setminus \frac{1}{4}$

Oct 22, 2017

$- \frac{1}{4}$

#### Explanation:

For those who have not learned L'Hopital's Rule, an alternative method relies upon some clever manipulations. First, combine the fractions in the numerator:

${\lim}_{x \to 2} \frac{\frac{1}{x} - \frac{1}{2}}{x - 2} = {\lim}_{x \to 2} \frac{\frac{2}{2 x} - \frac{x}{2 x}}{x - 2} = {\lim}_{x \to 2} \frac{\frac{2 - x}{2 x}}{x - 2}$

Now, factor out a -1 from the numerator of the combined fraction in the numerator, and then simplify the entire expression:

$= {\lim}_{x \to 2} \frac{\frac{- \left(x - 2\right)}{2 x}}{x - 2} = {\lim}_{x \to 2} \left(- \frac{\cancel{x - 2}}{2 x} \cdot \frac{1}{\cancel{x - 2}}\right)$

$= {\lim}_{x \to 2} \frac{- 1}{2 x} = - \frac{1}{4}$