First guess to multiply by $x^{-1.4}$ so the radical in numerator with $x^7$ becomes 1 and other stuff becomes 0. But then denominator becomes $-\infty$. What is the right approach?
$$ \lim_{x \to \infty} \frac{\sqrt[5]{x^7+3} – \sqrt[4]{2x^3 – 1}}{\sqrt[8]{x^7 + x^2 + 1} – x} $$
Best Answer
For starters: Note, that the denominator has as highest power of $x$ the $x^1$, so we multiply by $1 = \frac {1/x}{1/x}$, giving $$\frac{\sqrt[5]{x^7+3} - \sqrt[4]{2x^3 - 1}}{\sqrt[8]{x^7 + x^2 + 1} - x} = \frac{\sqrt[5]{x^2 + 3x^{-5}} - \sqrt[4]{2x^{-1} - x^{-4}}}{\sqrt[8]{x^{-1} + x^{-6} + x^{-8}} - 1} $$ Now the denominator converges to $-1$ for $x \to \infty$, and the numerator diverges to $\infty$, hence the fraction diverges to $-\infty$.