Question

35-38: Use continuity to evaluate the limit.

35. \(\mathop {\lim }\limits_{x \to 2} x\sqrt {20 - {x^2}} \)

Short answer

The value of the limit is 8.

Step-by-step solution

Theorems of continuity

When the functions say\(f\)and\(g\)arecontinuous at some number. Then, the following functions are also continuous:

1. \(f + g\)
2. \(f - g\)
3. \(cf\)
4. \(fg\)
5. \(\frac{f}{g},\,\,{\mathop{\rm if}\nolimits} \,\,g\left( a \right) \ne 0\)

Eachpolynomial iscontinuous on\(\mathbb{R} = \left( { - \infty ,\infty } \right)\). Anyrational functionis continuous in its domain.

When \(f\), and \(g\) are continuous at some number, then the composite function\(f \circ g\)is \(\left( {f \circ g} \right)\left( x \right) = f\left( {g\left( x \right)} \right)\) also continuous.

Use continuity to evaluate the limit

The product of function\(f\left( x \right) = x\sqrt {20 - {x^2}} \)is continuous on the interval\( - \sqrt {20} \le x \le \sqrt {20} \)since\(x\)is continuous on\(\mathbb{R}\)and\(\sqrt {20 - {x^2}} \)is continuous on its domain\( - \sqrt {20} \le x \le \sqrt {20} \).

It is observed that the number 2 is in that domain of\(f\). Therefore, the function\(f\)is continuous at 2.

\(\begin{aligned}\mathop {\lim }\limits_{x \to 2} f\left( x \right) &= f\left( 2 \right)\\ &= 2\sqrt {16} \\ &= 8\end{aligned}\)

Thus, the value of the limit is 8.