Chapter 1 - Limits and Their Properties - 1.5 Exercises - Page 89: 48

$\lim\limits_{x\to\frac{1}{2}^+}(x^2\tan{(\pi x)})=-\infty.$

$\lim\limits_{x\to\frac{1}{2}^+}(x^2\tan{(\pi x)})\to$ $\lim\limits_{x\to\frac{1}{2}^+}x^2=(\frac{1}{2}^+)^2=\dfrac{1}{4}.$ $\lim\limits_{x\to\frac{1}{2}^+}\tan{(\pi x)}=\tan{(\pi(\frac{1}{2}^+))}=-\infty.$ By Theorem $1.15:$ $\lim\limits_{x\to\frac{1}{2}^+}(x^2\tan{(\pi x)})=-\infty.$