Question

11-34: Evaluate the limit, if it exists.

34. \(\mathop {\lim }\limits_{h \to 0} \left( {\frac{{\frac{1}{{{{\left( {x + h} \right)}^2}}} - \frac{1}{{{x^2}}}}}{h}} \right)\)

Short answer

The solution of the limit is \( - \frac{2}{{{x^3}}}\).

Step-by-step solution

The limit laws

Let \(c\) be a constant and the limits \(\mathop {\lim }\limits_{x \to a} f\left( x \right)\) and \(\mathop {\lim }\limits_{x \to a} g\left( x \right)\) exists. Then

1. \(\mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) + g\left( x \right)} \right) = \mathop {\lim }\limits_{x \to a} f\left( x \right) + \mathop {\lim }\limits_{x \to a} g\left( x \right)\)
2. \(\mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - g\left( x \right)} \right) = \mathop {\lim }\limits_{x \to a} f\left( x \right) - \mathop {\lim }\limits_{x \to a} g\left( x \right)\)
3. \(\mathop {\lim }\limits_{x \to a} \left( {cf\left( x \right)} \right) = c\mathop {\lim }\limits_{x \to a} f\left( x \right)\)
4. \(\mathop {\lim }\limits_{x \to a} \left( {f\left( x \right)g\left( x \right)} \right) = \mathop {\lim }\limits_{x \to a} f\left( x \right) \cdot \mathop {\lim }\limits_{x \to a} g\left( x \right)\)
5. \(\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{{\mathop {\lim }\limits_{x \to a} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to a} g\left( x \right)}}\,\,{\mathop{\rm if}\nolimits} \,\,\mathop {\lim }\limits_{x \to a} g\left( x \right) \ne 0\).
6. \(\mathop {\lim }\limits_{x \to a} {\left( {f\left( x \right)} \right)^n} = {\left( {\,\mathop {\lim }\limits_{x \to a} f\left( x \right)} \right)^n}\), where \(n\) is a positive integer
7. \(\mathop {\lim }\limits_{x \to a} \sqrt[n]{{f\left( x \right)}} = \,\sqrt[n]{{\mathop {\lim }\limits_{x \to a} f\left( x \right)}}\), where \(n\) is a positive integer

(If \(n\) is even, we assume that \(\,\mathop {\lim }\limits_{x \to a} f\left( x \right) > 0\).)

1. \(\,\mathop {\lim }\limits_{x \to a} c = c\)
2. \(\,\mathop {\lim }\limits_{x \to a} x = a\)
3. \(\,\mathop {\lim }\limits_{x \to a} {x^n} = {a^n}\), where \(n\) is a positive integer
4. \(\,\mathop {\lim }\limits_{x \to a} \sqrt[n]{x} = \sqrt[n]{a}\), where \(n\) is a positive integer

(If \(n\) is even, we assume that \(a > 0\).)

 Step 2: Evaluate the limit

Evaluate the limit as shown below:

\(\begin{array}{c}\mathop {\lim }\limits_{h \to 0} \frac{{\frac{1}{{{{\left( {x + h} \right)}^2}}} - \frac{1}{{{x^2}}}}}{h} &=& \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{{x^2} - {{\left( {x + h} \right)}^2}}}{{{x^2}{{\left( {x + h} \right)}^2}}}}}{h}\\ &=& \mathop {\lim }\limits_{h \to 0} \frac{{{x^2} - \left( {{x^2} + 2xh + {h^2}} \right)}}{{h{x^2}{{\left( {x + h} \right)}^2}}}\\ &=& \mathop {\lim }\limits_{h \to 0} \frac{{{x^2} - \left( {{x^2} + 2xh + {h^2}} \right)}}{{h{x^2}{{\left( {x + h} \right)}^2}}}\end{array}\)

Solve further,

\(\begin{array}{c}\mathop {\lim }\limits_{h \to 0} \frac{{\frac{1}{{{{\left( {x + h} \right)}^2}}} - \frac{1}{{{x^2}}}}}{h} &=& \mathop {\lim }\limits_{h \to 0} \frac{{ - 2xh - h}}{{h{x^2}{{\left( {x + h} \right)}^2}}}\\ &=& \mathop {\lim }\limits_{h \to 0} \frac{{ - h\left( {2x + h} \right)}}{{h{x^2}{{\left( {x + h} \right)}^2}}}\\ &=& \mathop {\lim }\limits_{h \to 0} \frac{{ - \left( {2x + h} \right)}}{{{x^2}{{\left( {x + h} \right)}^2}}}\\ &=& \frac{{ - 2x}}{{{x^2} \cdot {x^2}}}\\ &=& \frac{{ - 2}}{{{x^3}}}\end{array}\)

Thus, the solution of the limit is \( - \frac{2}{{{x^3}}}\).