Question

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

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

Short answer

The solution of the limit is \( - \frac{1}{4}\).

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 \(a > 0\).)

Evaluate the limit

Evaluate the given limit as shown below:

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

Solve further,

\(\begin{array}{c}\mathop {\lim }\limits_{h \to 0} \frac{{{{\left( { - 2 + h} \right)}^{ - 1}} + {2^{ - 1}}}}{h} &=& \mathop {\lim }\limits_{h \to 0} \frac{1}{{2\left( {h - 2} \right)}}\\ &=& \frac{1}{{2\left( {0 - 2} \right)}}\\ &=& - \frac{1}{4}\end{array}\)

Thus, the solution of the limit is \( - \frac{1}{4}\).