Question

Evaluate the limit if it exists.

\(\mathop {{\bf{lim}}}\limits_{h \to {\bf{0}}} \frac{{\sqrt {{\bf{9}} + h} - {\bf{3}}}}{h}\)

Short answer

The value of the limit is \(\frac{1}{6}\).

Step-by-step solution

Simplify the limit using factorization

Rationalize the expression \(\mathop {\lim }\limits_{h \to 0} \frac{{\sqrt {9 + h} - 3}}{h}\).

\(\begin{aligned}\mathop {\lim }\limits_{h \to 0} \frac{{\sqrt {9 + h} - 3}}{h} &= \mathop {\lim }\limits_{h \to 0} \frac{{\sqrt {9 + h} - 3}}{h} * \frac{{\sqrt {9 + h} + 3}}{{\sqrt {9 + h} + 3}}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {\sqrt {9 + h} } \right)}^2} - {{\left( 3 \right)}^2}}}{{h\left( {\sqrt {9 + h} + 3} \right)}}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{h}{{h\left( {\sqrt {9 + h} + 3} \right)}}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{1}{{\sqrt {9 + h} + 3}}\end{aligned}\)

Evaluate the limit

For the expression \(\mathop {\lim }\limits_{h \to 0} \frac{1}{{\sqrt {9 + h} + 3}}\), Quotient law is applicable.

According to the Quotient law, \(\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{f\left( x \right)}} = \frac{{\mathop {\lim }\limits_{x \to a} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to a} g\left( x \right)}}\), where \(\mathop {\lim }\limits_{x \to a} f\left( x \right)\) and \(\mathop {\lim }\limits_{x \to a} g\left( x \right)\) exists and \(\mathop {\lim }\limits_{x \to a} g\left( x \right) \ne 0\).

Solve the expression \(\mathop {\lim }\limits_{h \to 0} \frac{1}{{\sqrt {9 + h} + 3}}\).

\(\begin{aligned}\mathop {\lim }\limits_{h \to 0} \frac{1}{{\sqrt {9 + h} + 3}} &= \frac{{\mathop {\lim }\limits_{h \to 0} \left( 1 \right)}}{{\mathop {\lim }\limits_{h \to 0} \left( {\sqrt {9 + h} } \right) + \mathop {\lim }\limits_{h \to 0} \left( 3 \right)}}\\ &= \frac{1}{{\sqrt {9 + 0} + 3}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\rm{root law}}} \right)\\ &= \frac{1}{6}\end{aligned}\)

Thus, the value of the limit is \(\frac{1}{6}\).