Question

Evaluate the limit, if it exists.

\(\mathop {{\bf{lim}}}\limits_{h \to {\bf{0}}} \frac{{{{\left( {h - {\bf{3}}} \right)}^{\bf{2}}} - {\bf{9}}}}{h}\)

Short answer

The value of the limit is \( - 6\).

Step-by-step solution

Simplify the limit by using the factorization

Solve the numerator of the expression \(\mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {h - 3} \right)}^2} - 9}}{h}\).

\(\begin{aligned}\mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {h - 3} \right)}^2} - 9}}{h} &= \mathop {\lim }\limits_{h \to 0} \frac{{{h^2} + 9 - 6h - 9}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{{h^2} - 6h}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \left( {h - 6} \right)\end{aligned}\)

Evaluate the limit

For the expression \(\mathop {\lim }\limits_{h \to 0} \left( {h - 6} \right)\), Difference law is applicable.

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

Solve the expression \(\mathop {\lim }\limits_{h \to 0} \left( {h - 6} \right)\).

\(\begin{aligned}\mathop {\lim }\limits_{h \to 0} \left( {h - 6} \right) &= \mathop {\lim }\limits_{h \to 0} \left( h \right) - \mathop {\lim }\limits_{h \to 0} \left( 6 \right)\\ &= 0 - 6\\ &= - 6\end{aligned}\)

Thus, the value of the limit is \( - 6\).