What will be the value of this limit?

Question

I need help solving this limit.
$$ \large \lim_{x\to \infty} ([(x+a)(x+b)(x+c)]^{\frac 13} -x) $$

I have tried rationalizing and then dividing the numerator and the denominator by $x^3$ but that doesn't work. I still get a $\frac 00$ form.

All help will be appreciated

Answer 1

Short answer:

Applying the identity

$$\sqrt[3]u-v=\frac{u-v^3}{(\sqrt[3]u)^2+\sqrt[3]u\,v+v^2},$$

you understand that the numerator will be a polynomial with leading term $(a+b+c)x^2$, and the denominator will be a sum of three terms each asymptotic to $x^2$, hence

$$\frac{a+b+c}3.$$

Answer 2

HINT: use that $$a^3-b^3=(a-b)(a^2+ab+b^2)$$

Answer 3

Let $$f(x)=x^n+rx^{n-1}+\cdots$$ be a monic polynomial of degree $n$. Then $$f(x)^{1/n} =x(1+rx^{-1}+O(x^{-2}))^{1/n} =x(1+(r/n)x^{-1}+O(x^{-2}))$$ and so $$\lim_{x\to\infty}(f(x)^{1/n}-x)=\frac rn.$$

In your example, $f(x)=(x+a)(x+b)(x+c)$.

Answer 4

$$\large \lim_{x\to \infty} ([(x+a)(x+b)(x+c)]^{\frac 13} -x)$$ $$=\lim_{x\to \infty} (x(1+\frac{a}{x})^{\frac 13}(1+\frac{b}{x})^{\frac 13}(1+\frac{c}{x})^{\frac 13} -x)$$ $$=\lim_{x\to \infty} (x(1+\frac{a}{3x}+O(1/x^2))(1+\frac{b}{3x}+O(1/x^2))(1+\frac{c}{3x}+O(1/x^2)) -x)$$ $$=\lim_{x\to \infty} (x(1+\frac{a+b+c}{3x}+O(1/x^2))-x)$$ $$=\lim_{x\to \infty} \frac{a+b+c}{3}+O(\frac{1}{x})$$ $$=\frac{a+b+c}{3}$$