Question

Continuous Functions In Exercises 73 and \(74,\) find the value of \(c\) that makes the function continuous at \(x=0\). \(f(x)=\left\\{\begin{array}{ll}\frac{4 x-2 \sin 2 x}{2 x^{3}}, & x \neq 0 \\\ c, & x=0\end{array}\right.\)

Short answer

The constant \(c\) that makes the function continuous at \(x = 0\) is \(c = 2\).

Step-by-step solution

Calculate the Limit of \(f(x)\) as \(x\) Approaches Zero

The first step is to calculate \(\lim_{{x \to 0}} f(x) = \lim_{{x \to 0}} \frac{4x - 2\sin{2x}}{2x^3}\). Using L'Hopital's rule, which states that the limit of a ratio of two functions is equal to the limit of the ratios of their derivatives, we get \(\lim_{{x \to 0}} \frac{4 - 4\cos{2x}}{6x^2}\). Again applying L'Hopital's rule, this becomes \(\lim_{{x \to 0}} \frac{8\sin{2x}}{12x}\), which simplifies further to \(\lim_{{x \to 0}} \frac{\sin{2x}}{x} = 2\) using the standard limit \(\lim_{{x \to 0}} \frac{\sin{x}}{x} = 1\).

Equate the Limit of \(f(x)\) as \(x\) Approaches Zero to \(f(0)\)

To ensure the function is continuous at \(x = 0\), its value at that point must match the limit as \(x\) approaches zero. From step 1, we know \(\lim_{{x \to 0}} f(x) = 2\). The function is defined to be equal to \(c\) at \(x = 0\), therefore, to make the function continuous at \(x = 0\), \(c\) must be equal to 2.

Conclusion

After analyzing the limit of the function as \(x\) approaches zero and the function value at \(x = 0\), it is concluded that the function is continuous at \(x = 0\) when \(c = 2\). This is the value that makes the function continuous at \(x = 0\).