Find the limit -> Infinity with radicals 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?
$$ \lim_{x \to \infty} \frac{\sqrt[5]{x^7+3} - \sqrt[4]{2x^3 - 1}}{\sqrt[8]{x^7 + x^2 + 1} - x} $$
 A: 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 $-\infty$.
A: Putting $\frac1x=h$
$$\sqrt[5]{x^7+3} - \sqrt[4]{2x^3 - 1}=\frac{(1+3h^7)^{\frac15}}{h^{\frac75}}-\frac{(2-h^3)^{\frac14}}{h^{\frac34}}=\frac{(1+3h^7)^{\frac15}-h^{\frac{13}{20}}(2-h^3)^{\frac14}}{h^{\frac75}}$$ as $\frac75-\frac34=\frac{7\cdot4-5\cdot3}{20}=\frac{13}{20}$
$$\sqrt[8]{x^7 + x^2 + 1} - x=\frac{(1+h^5+h^7)^{\frac18}}{h^{\frac78}}-\frac1h=\frac{h^{\frac18}(1+h^5+h^7)^{\frac18}-1}h$$
as $1-\frac78=\frac18$
$$\implies\lim_{x \to \infty} \frac{\sqrt[5]{x^7+3} - \sqrt[4]{2x^3 - 1}}{\sqrt[8]{x^7 + x^2 + 1} - x}=\lim_{h\to0} \frac{(1+3h^7)^{\frac15}-h^{\frac{13}{20}}(2-h^3)^{\frac14}}{h^{\frac75}}\cdot \frac h{h^{\frac18}(1+h^5+h^7)^{\frac18}-1}$$
$$=\lim_{h\to0} \frac{(1+3h^7)^{\frac15}-h^{\frac{13}{20}}(2-h^3)^{\frac14}}{h^{\frac25}}\cdot \frac 1{h^{\frac18}(1+h^5+h^7)^{\frac18}-1}\left(\text{ as }\frac75-1=\frac25\right)$$
$$=\frac10\cdot\frac1{-1}=-\infty$$
