Question

Evaluate the limit, if it exists.

\(\mathop {{\bf{lim}}}\limits_{x \to {\bf{9}}} \frac{{{\bf{9}} - x}}{{{\bf{3}} - \sqrt x }}\)

Short answer

The value of the limits is 6.

Step-by-step solution

Simplify the expression \(\mathop {{\bf{lim}}}\limits_{x \to {\bf{9}}} \frac{{{\bf{9}} - x}}{{{\bf{3}} - \sqrt x }}\)

Multiply and divide the fraction by \(3 + \sqrt x \) in the expression \(\mathop {\lim }\limits_{x \to 9} \frac{{9 - x}}{{3 - \sqrt x }}\).

\(\begin{aligned}\mathop {\lim }\limits_{x \to 9} \frac{{9 - x}}{{3 - \sqrt x }} &= \mathop {\lim }\limits_{x \to 9} \frac{{9 - x}}{{3 - \sqrt x }} * \frac{{3 + \sqrt x }}{{3 + \sqrt x }}\\ &= \mathop {\lim }\limits_{x \to 9} \frac{{\left( {9 - x} \right)\left( {3 + \sqrt x } \right)}}{{9 - x}}\\ &= \mathop {\lim }\limits_{x \to 9} \left( {3 + \sqrt x } \right)\end{aligned}\)

Evaluate the limit

For the expression,\(\mathop {\lim }\limits_{x \to 9} \left( {3 + \sqrt x } \right)\) Difference law is applicable.

According to the sum 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 \(\mathop {\lim }\limits_{x \to 9} \left( {3 + \sqrt x } \right)\).

\(\begin{aligned}\mathop {\lim }\limits_{x \to 9} \left( {3 + \sqrt x } \right) &= \mathop {\lim }\limits_{x \to 9} \left( 3 \right) + \mathop {\lim }\limits_{x \to 9} \left( {\sqrt x } \right)\\ &= 3 + \sqrt {\mathop {\lim }\limits_{x \to 9} \left( x \right)} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\rm{Root}}\;{\rm{law}}} \right)\\ &= 3 + \sqrt 9 \\ &= 6\end{aligned}\)

Thus, the value of the limit is 6.