AP Calculus AB and BC

Limit of (1/x-1/2)/(x-2) at x = 2 Is -1/4

Q: Where exactly does the minus sign come from?

From 2 - x = -(x - 2). Combining \frac{1}{x}-\frac{1}{2} over 2x puts 2 - x on top, while the outer denominator carries x - 2. Those are negatives of each other, so cancelling them leaves a factor of -1 behind.

Q: Is there a way to see the answer without the algebra?

Yes. The expression is the difference quotient for f(t) = \frac{1}{t} at t = 2, so the limit is f'(2) = -\frac{1}{2^2} = -\frac{1}{4}. Anything shaped like \frac{f(x)-f(a)}{x-a} can be read off as f'(a) whenever that derivative exists.

Q: Does L'Hopital's rule work here?

It does. The form is \frac{0}{0}, the numerator differentiates to -\frac{1}{x^2} because the constant \frac{1}{2} drops out, the denominator differentiates to 1, and -\frac{1}{x^2} is -\frac{1}{4} at x = 2. Unit 1 expects the algebra, since the rule arrives later in the course.

The limit of (1/x - 1/2)/(x - 2) as x approaches 2 is -1/4. Direct substitution gives 0/0. Combine the numerator over the common denominator 2x to get (2 - x)/(2x), then use 2 - x = -(x - 2) to cancel. That leaves -1/(2x), which is -1/4 at x = 2. The minus sign is the difficulty.

\lim_{x \to 2} \frac{\frac{1}{x}-\frac{1}{2}}{x-2} = -\frac{1}{4}

Settled by clearing the complex fraction.

Combining the numerator, then cancelling

Deal with the small fractions first. The two terms upstairs share the common denominator $2x$ .

\frac{1}{x}-\frac{1}{2} = \frac{2}{2x}-\frac{x}{2x} = \frac{2-x}{2x}

Dividing by $x - 2$ is multiplying by $\frac{1}{x-2}$ , so the compound fraction flattens into a single one.

\frac{\frac{1}{x}-\frac{1}{2}}{x-2} = \frac{2-x}{2x}\cdot\frac{1}{x-2} = \frac{2-x}{2x(x-2)}

Now the step that decides the sign. The numerator $2 - x$ is the negative of the factor $x - 2$ sitting downstairs, so pulling out a $-1$ makes the cancellation visible.

\frac{2-x}{2x(x-2)} = \frac{-(x-2)}{2x(x-2)} = -\frac{1}{2x}

What is left is continuous at $2$ , so substitution finishes the job.

\lim_{x \to 2} \left(-\frac{1}{2x}\right) = -\frac{1}{2 \cdot 2} = -\frac{1}{4}

What direct substitution gives

At $x = 2$ the numerator is $\frac{1}{2}-\frac{1}{2} = 0$ and the denominator is $2 - 2 = 0$ .

\frac{\frac{1}{2}-\frac{1}{2}}{2-2} = \frac{0}{0}

Indeterminate, and the stacked shape hides the shared factor. Combining the two small fractions is what brings $x - 2$ into the open where it can cancel, which is why the first move is arithmetic rather than calculus.

A sign check with numbers

The function is negative on both sides of $2$ : about $-0.24988$ at $x = 2.001$ and about $-0.25013$ at $x = 1.999$ . An answer of $+\frac{1}{4}$ contradicts the arithmetic before it contradicts the algebra.

The derivative hiding in the problem

Set $f(t) = \frac{1}{t}$ . Then $f(2) = \frac{1}{2}$ , so the whole expression is $\frac{f(x)-f(2)}{x-2}$ , the difference quotient for $f$ based at $2$ .

\lim_{x \to 2} \frac{\frac{1}{x}-\frac{1}{2}}{x-2} = f'(2), \qquad f(t) = \frac{1}{t}

The power rule gives $f'(t) = -\frac{1}{t^2}$ , so $f'(2) = -\frac{1}{4}$ . Read this way, the sign needs no algebra at all: $\frac{1}{x}$ is decreasing, so its rate of change is negative.

The mistakes students make

Answering $\frac{1}{4}$ . Losing the minus sign in $2 - x = -(x-2)$ is the single most common error on this problem, and the sample values above rule that answer out.
Using $x - 2$ as the common denominator for $\frac{1}{x} - \frac{1}{2}$ . The small fractions need $2x$ , while the outer $x - 2$ belongs to the division.
Multiplying by $x - 2$ instead of dividing. A fraction divided by $x - 2$ is a fraction times $\frac{1}{x-2}$ .
Reporting that the limit does not exist. The function has no value at $x = 2$ , but the discontinuity there is removable and the limit is finite.

Not sure which technique a limit wants?

The Limit Method Chooser walks the decision from direct substitution through factoring, the conjugate, and L'Hopital, and says why each one applies or fails.

Frequently asked questions

Where exactly does the minus sign come from?

From $2 - x = -(x - 2)$ . Combining $\frac{1}{x}-\frac{1}{2}$ over $2x$ puts $2 - x$ on top, while the outer denominator carries $x - 2$ . Those are negatives of each other, so cancelling them leaves a factor of $-1$ behind.

Is there a way to see the answer without the algebra?

Yes. The expression is the difference quotient for $f(t) = \frac{1}{t}$ at $t = 2$ , so the limit is $f'(2) = -\frac{1}{2^2} = -\frac{1}{4}$ . Anything shaped like $\frac{f(x)-f(a)}{x-a}$ can be read off as $f'(a)$ whenever that derivative exists.

Does L'Hopital's rule work here?

It does. The form is $\frac{0}{0}$ , the numerator differentiates to $-\frac{1}{x^2}$ because the constant $\frac{1}{2}$ drops out, the denominator differentiates to $1$ , and $-\frac{1}{x^2}$ is $-\frac{1}{4}$ at $x = 2$ . Unit 1 expects the algebra, since the rule arrives later in the course.