Problem 32 - Precalculus with Limits

Question

In Exercises 9-36, find the limit (if it exists). Use a graphing utility to verify your result graphically. $$\lim_{x \to 0} \dfrac{\dfrac{1}{2+x}-\dfrac{1}{2}}{x}$$

Step-by-Step Solution

Answer

-1/2

Step 1: Simplify the Equation

To simplify this equation, we will use the property \(\dfrac{a}{c} - \dfrac{b}{c} = \dfrac{a - b}{c}\) to combine the two fractions into one: \[\lim_{x \to 0} \dfrac{\dfrac{1}{2+x}-\dfrac{1}{2}}{x} = \lim_{x \to 0} \dfrac{\dfrac{2}{2+x}-1}{x} = \lim_{x \to 0} \dfrac{2 - (2+x)}{x(2+x)}\]

Step 2: Substitute \(x\) for 0

Now, substitute \(x\) for 0: \[\lim_{x \to 0} \dfrac{2 - 2}{0(2+0)} = \dfrac{0}{0}\] Since we have an indeterminate form 0/0, we can't find the limit directly.

Step 3: Apply L'Hopital's Rule

To find the limit \( \lim_{x \to 0} \dfrac{0}{0}\), apply L'Hopital's Rule to calculate the limit by computing the derivatives of the numerator and the denominator. The derivative of the numerator \((2 - 2 - x)\) is -1. The derivative of the denominator \((x(2+x))\) is \(2+x\). So the new limit to calculate is: \[\lim_{x \to 0} \dfrac{-1}{2+x}\]

Step 4: Evaluate the Limit

Now we can substitute \(x\) for 0 in the equation: \[\lim_{x \to 0} \dfrac{-1}{2+0} = -\dfrac{1}{2}\] So \(\lim_{x \to 0} \dfrac{\dfrac{1}{2+x}-\dfrac{1}{2}}{x} = -\dfrac{1}{2}\)