$\lim_{x\rightarrow +\infty}{8x(x+1)-\sqrt{4x^2+2x}\cdot\sqrt[3]{64x^3+144x^2+90x+17}}$

Question

$$\lim_{x\rightarrow +\infty}{8x(x+1)-\sqrt{4x^2+2x}\cdot\sqrt[3]{64x^3+144x^2+90x+17}}$$

I know that it's $\infty - \infty$, but the square root and cube root make it too complicated. Can anyone help me to deal with this problem? Thank you

Answer 1

Hint

First, notice that $$64x^3+144x^2+90x+17=(2 x+1) \left(32 x^2+56 x+17\right)$$

One of my professors used to tell us "We are almost closer to $0$ than to $\infty$". So, start making $x=\frac 1 y$ and simplify to make the expression $$\frac{8 (1+y)-\sqrt{2} (2+y)^{5/6} \sqrt[3]{ 32+56y+17 y^2}}{y^2}$$ So, for the limit, we just need to expand the numerator to $O(y^3)$ which is not much work using the binomial expansion.