Answer
(a)$$\lim_{x \to 0}\frac{\tan 2x}{x}=1$$
(b) Yes, by defining $f(0)=1$.
Work Step by Step
(a)
As we know,$$\tan2x=\frac{\sin 2x}{\cos 2x}=\frac{2 \sin x \cos x}{\cos ^2 x - \sin ^2 x}.$$ So we have$$\lim_{x \to 0}\frac{\tan 2x}{x}=\lim_{x \to 0}\frac{\frac{2 \sin x \cos x}{\cos ^2 x- \sin ^2 x}}{x}=\lim_{x\to 0}\left ( \frac{\sin x}{x} \right ) \left ( \frac{2 \cos x}{\cos ^2 x - \sin ^2 x} \right )= (1)(1)=1$$(Please note that in finding the limit we have used Theorem 1.9 (1), $\lim_{x \to 0}\frac{\sin x}{x}=1$).
(b)
Since $x=0$ is a removable discontinuity for the function $f(x)$ (because $\lim_{x \to 0}f(x)$ does exist), we can make the function $f(x)$ continuous by defining its value at $x=0$ to be its limit at $x=0$; that is,$$f(0)=\lim_{x \to 0} f(x)=1.$$
