Question

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

Answer 1

`-\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:

`\lim_{x \rightarrow 2} \frac { \frac {1} {x} - \frac {1} {2} } {x - 2}=\lim_{x \rightarrow 2} \frac { -\frac {1} {x^2} - 0} {1 - 0}=\lim_{x \rightarrow 2} (-\frac {1} {x^2})=-\frac {1} {2^2}=-\frac{1}{4}`

Answer 2

`-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->2} (1/x - 1/2)/(x-2) = lim_{x->2} (2/(2x) - x/(2x))/(x-2) = lim_{x->2} ((2-x)/(2x))/(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->2} ((-(x-2))/(2x))/(x-2) = lim_{x->2} (-(cancel(x-2))/(2x)*1/cancel(x-2))`

`= lim_{x->2} (-1)/(2x) = -1/4`