Question

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

29. \(\mathop {\lim }\limits_{x \to 16} \frac{{4 - \sqrt x }}{{16x - {x^{\bf{2}}}}}\)

Short answer

The solution of the limit is \(\frac{1}{{128}}\).

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

Quotient law cannot be applied since the limit of the denominator is 0.

So, rationalize the numerator and evaluate the limit as shown below:

\(\begin{array}{c}\mathop {\lim }\limits_{x \to 16} \frac{{4 - \sqrt x }}{{16x - {x^2}}} &=& \mathop {\lim }\limits_{x \to 16} \frac{{4 - \sqrt x }}{{16x - {x^2}}} \cdot \frac{{4 + \sqrt x }}{{4 + \sqrt x }}\\ &=& \mathop {\lim }\limits_{x \to 16} \frac{{\left( {4 - \sqrt x } \right) \cdot \left( {4 + \sqrt x } \right)}}{{\left( {16x - {x^2}} \right)\left( {4 + \sqrt x } \right)}}\\ &=& \mathop {\lim }\limits_{x \to 16} \frac{{16 - x}}{{x\left( {16 - x} \right)\left( {4 + \sqrt x } \right)}}\end{array}\)

Solve further,

\(\begin{array}{c}\mathop {\lim }\limits_{x \to 16} \frac{{4 - \sqrt x }}{{16x - {x^2}}} &=& \mathop {\lim }\limits_{x \to 16} \frac{1}{{x\left( {4 + \sqrt x } \right)}}\\ &=& \frac{1}{{16\left( {4 + \sqrt {16} } \right)}}\\ &=& \frac{1}{{16\left( 8 \right)}}\\ &=& \frac{1}{{128}}\end{array}\)

Thus, the solution of the limit is \(\frac{1}{{128}}\).