$\underset{x→-∞}{\lim}\frac{x^5\tan\left(\frac{1}{πx^2}\right)+3|x|^2+7}{|x|^3+7|x|+8}$ is equal to

$-\frac{1}{π}$

$\underset{x→-∞}{\lim}\frac{x^5\tan\left(\frac{1}{πx^2}\right)+3|x|^2+7}{|x|^3+7|x|+8}=\underset{x→-∞}{\lim}\frac{x^5\tan\left(\frac{1}{πx^2}\right)+3x^2+7}{-x^3-7x+8}$  $∵ x < 0⇒|x|=-x$

$\underset{x→-∞}{\lim}\frac{x^2\tan\left(\frac{1}{πx^2}\right)+3/x+7/x^3}{-1-7/x^2+8/x^3}$ (Dividing the numerator and denominator by $x^3$)

$\underset{x→-∞}{\lim}\frac{\frac{1}{π}.\frac{\tan\{1/(πx^2)\}}{1/(πx^2)}+\frac{3}{x}+\frac{7}{x^3}}{-1-\frac{7}{x^2}+\frac{8}{x^3}}=\frac{\frac{1}{π}.1+0+0}{-1-0+0}=-\frac{1}{π}$

$\left[∵x→∞=⇒\frac{1}{πx^2}→0\,and\,\underset{x→0}{\lim}\frac{\tan θ}{θ}\right]$