Question

35-38: Use continuity to evaluate the limit.

38. \(\mathop {\lim }\limits_{x \to 4} {3^{\sqrt {{x^2} - 2x - 4} }}\)

Short answer

The solution of the limit is 9.

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. \(\frac{f}{g},\,\,{\mathop{\rm if}\nolimits} \,\,g\left( a \right) \ne 0\)

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

When\(f\), and\(g\)are continuous at some number, then thecomposite 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

It is observed that the function\(f\left( x \right) = {3^{\sqrt {{x^2} - 2x - 4} }}\)is within an exponential function and a root function.

Therefore, the function\(f\left( x \right)\)is continuous throughout its domain.

The domain of the function is shown below:

\(\begin{aligned}\left\{ {x|{x^2} - 2x - 4 \ge 0} \right\} &= \left\{ {x|{x^2} - 2x + 1 \ge 5} \right\}\\ &= \left\{ {x|{{\left( {x - 1} \right)}^2} \ge 5} \right\}\\ &= \left\{ {x|\left| {x - 1} \right| \ge \sqrt 5 } \right\}\\ &= \left( { - \infty ,1 - \sqrt 5 } \right) \cup \left( {1 + \sqrt 5 ,\infty } \right)\end{aligned}\)

It is observed that the number 4 is in that domain. Hence, the function\(f\)is continuous at 4.

Evaluate the limit as shown below:

\(\begin{aligned}\mathop {\lim }\limits_{x \to 4} f\left( x \right) &= f\left( 4 \right)\\ &= {3^{\sqrt {16 - 8 - 4} }}\\ &= {3^2}\\ &= 9\end{aligned}\)

Thus, the solution of the limit is 9.