Solveeit Logo
Login

Question

Question: $\lim_{x \to -\infty} \frac{x^5 \tan(\frac{1}{\pi x^2})+3|x|^2+7}{|x|^3+7|x|+8}$...

lim⁡x→−∞x5tan⁡(1πx2)+3∣x∣2+7∣x∣3+7∣x∣+8\lim_{x \to -\infty} \frac{x^5 \tan(\frac{1}{\pi x^2})+3|x|^2+7}{|x|^3+7|x|+8}

A

= −1π\frac{-1}{\pi}

B

= π\pi

C

= 1π\frac{1}{\pi}

D

Does not exist

Answer

= −1π\frac{-1}{\pi}

Explanation

Solution

For x→−∞x \to -\infty, ∣x∣=−x|x| = -x.

lim⁡x→−∞x5tan⁡(1πx2)+3x2+7−x3−7x+8\lim_{x \to -\infty} \frac{x^5 \tan(\frac{1}{\pi x^2})+3x^2+7}{-x^3-7x+8}

As x→−∞x \to -\infty, 1πx2→0\frac{1}{\pi x^2} \to 0. Using tan⁡y≈y\tan y \approx y for small yy:

x5tan⁡(1πx2)≈x5⋅1πx2=x3πx^5 \tan(\frac{1}{\pi x^2}) \approx x^5 \cdot \frac{1}{\pi x^2} = \frac{x^3}{\pi}.

The limit becomes lim⁡x→−∞x3π+3x2+7−x3−7x+8\lim_{x \to -\infty} \frac{\frac{x^3}{\pi}+3x^2+7}{-x^3-7x+8}.

Divide numerator and denominator by x3x^3:

lim⁡x→−∞1π+3x+7x3−1−7x2+8x3=1π+0+0−1−0+0=−1π\lim_{x \to -\infty} \frac{\frac{1}{\pi}+\frac{3}{x}+\frac{7}{x^3}}{-1-\frac{7}{x^2}+\frac{8}{x^3}} = \frac{\frac{1}{\pi}+0+0}{-1-0+0} = -\frac{1}{\pi}.