Question

Show that the function

\(f\left( x \right) = \left\{ \begin{array}{l}{x^4}\sin \left( {\frac{1}{x}} \right)\,\,\,\,\,{\mathop{\rm if}\nolimits} \,\,x\,\,{\mathop{\rm is}\nolimits} \,\,\,{\mathop{\rm rational}\nolimits} \\0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\mathop{\rm if}\nolimits} \,\,x\,\,{\mathop{\rm is}\nolimits} \,\,\,{\mathop{\rm irrational}\nolimits} \end{array} \right.\)is continuous on \(\left( { - \infty ,\infty } \right)\).

Short answer

It is proved that the function \(f\) is continuous on \(\left( { - \infty ,\infty } \right)\).

Step-by-step solution

Theorem of continuity

Theorem 7states that the types of functions arecontinuousat all numbers in their domains are as follows:

• Polynomials
• Rational functions
• Root function
• Trigonometric functions
• Inverse trigonometric functions
• Exponential functions
• Logarithmic functions.

Theorem 9states that when\(g\)is continuous at\(a\)and\(f\)is continuous at\(g\left( a \right)\), then thecomposite function\(f \circ g\)is given by \(\left( {f \circ g} \right)\left( x \right) = f\left( {g\left( x \right)} \right)\) is continuous at \(a\).

Step 2:Show that the function \(f\left( x \right)\) is continuous on \(\left( { - \infty ,\infty } \right)\)

The Squeeze theoremstates that when\(f\left( x \right) \le g\left( x \right) \le h\left( x \right)\)if\(x\)is near\(a\)(except possibly at\(a\)).

\(\mathop {\lim }\limits_{x \to a} f\left( x \right) = \mathop {\lim }\limits_{x \to a} h\left( x \right) = L\)then\(\mathop {\lim }\limits_{x \to a} g\left( x \right) = L\)

It is observed that the function is the product of a polynomial and composite of a trigonometric function and a rational function. Therefore, the function\(f\left( x \right) = {x^4}\sin \left( {\frac{1}{x}} \right)\)is continuous on\(\left( { - \infty ,0} \right) \cup \left( {0,\infty } \right)\). Then

\( - 1 \le \sin \left( {\frac{1}{x}} \right) \le 1\)

Multiply by\({x^4}\)in the above inequality as shown below:

\( - {x^4} \le {x^4}\sin \left( {\frac{1}{x}} \right) \le {x^4}\)

The Squeeze theorem provides us\(\mathop {\lim }\limits_{x \to 0} \left( {{x^4}\sin \left( {\frac{1}{x}} \right)} \right) = 0\)that is,\(f\left( 0 \right)\)since\(\mathop {\lim }\limits_{x \to 0} \left( { - {x^4}} \right) = 0\)and\(\mathop {\lim }\limits_{x \to 0} {x^4} = 0\).

This means that the function\(f\)is continuous at 0 and therefore, on\(\left( { - \infty ,\infty } \right)\).

Thus, it is proved that the function \(f\) is continuous on \(\left( { - \infty ,\infty } \right)\).