Question

lim \(\dfrac{-2x^5+x^4-3}{3x^2-7}\)

x-> -∞

Answer 1

\(\lim\limits_{x\rightarrow-\infty}\dfrac{-2x^5+x^4-3}{3x^2-7}=\lim\limits_{x\rightarrow-\infty}\dfrac{x^5\left(-2+\dfrac{1}{x}-\dfrac{3}{x^5}\right)}{x^2\left(3-\dfrac{7}{x^2}\right)}\\ =\lim\limits_{x\rightarrow-\infty}\dfrac{x^3\left(-2+\dfrac{1}{x}-\dfrac{3}{x^5}\right)}{3-\dfrac{7}{x^2}}=+\infty\)

Answer 2

\(= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{ - 2 + \dfrac{1}{x} - \dfrac{3}{{{x^5}}}}}{{\dfrac{3}{{{x^3}}} - \dfrac{7}{{{x^5}}}}} = \dfrac{{ - 2}}{{{0^ - }}} = + \infty \)